Progressions-A.P. and G.P. (Part-5)

© ® Written by Sankar Ghosh
 
Dear Friends,

In my blog series “Progressions – A.P. and G.P.”, the first three parts were dedicated to Arithmetic Progression (A.P.), where I explained the concepts in detail along with examples and exercises.
In Part 4, we moved into Geometric Progression (G.P.), focusing on the general term and solving related problems.

I encourage you to try the exercises given at the end of each part—your learning will be complete only when you practice them thoroughly. And of course, if you face any difficulty or have doubts, please feel free to write in the comments. I will be more than happy to help you out.

Now, in today’s post, we take the next step in our journey with G.P. — discussing the sum to n terms of a Geometric Progression.

Happy learning and keep sharing your thoughts and questions with me!

Warm regards,
Sankar Ghosh



Sum of the first `n` terms of a G.P.

Let the first term of a geometric progression be `a`, common ratio `r\left(\ne1\right)` and sum of first `n` terms is `S_n`

`\therefore`        `S_n=a+ar+ar^2+ar^3+...................+ar^{n-2}+ar^{n-1}`
and     `rS_n=        ar+ar^2+ar^3+ar^4+...........+ar^{n-2}+ar^{n-1}+ar^n`

.......................................................................................................................................

(Subtracting)    `S_n-rS_n=a-ar^n=a\left(1-r^n\right)`
`\therefore`      `S_n=\frac{a\left(1-r^n\right)}{1-r}\or\frac{a\left(r^n-1\right)}{r-1}`.

Thus, we can determine the sum of first `n` terms of a G.P. if we have the first term and common ratio of that G.P.

Note (1): If `r=1` then `S_n=a+a+a+a+...+` upto `n` terms `=na`.

         (2): `S_n=t_1+t_2+t_3+.......+t_n`  and `S_{n-1}=t_1+t_2+t_3+......+t_{n-1}`.
               `\therefore t_n=S_n-S_{n-1}`. 

        (3): If the last term of a G.P. be `l` then `S_n=a\left(\frac{r^n-1}{r-1}\right)=\frac{ar^n-a}{r-1}=\frac{\left(ar^{n-1\right)}r-a}{r-1}`

`\implies`        `S_n=\frac{lr-a}{r-1}\or\frac{a-lr}{1-r},r\ne1`  where `l=ar^{n-1}`.


ILLUSTRATIVE EXAMPLES


  1. Find the sum of `8` terms of the G.P. `1,3,9,27...` 
    Solution:     The first term and common ratio of the given G.P. are `a=1,r=3` respectively.`\therefore` sum to first `8` terms of the given G.P. `S_8=\frac{a\left(r^8-1\right)}{r-1}=\frac{3^8-1}{3-1}=\frac{6561-1}{3-1}=\frac{6560}{2}=3280`.
  2. How many terms of the given geometric series `2+4+8+16+.....`will make the sum `2046?
    Solution: Here, the first term `a=2` and common ratio `\frac{4}{2}=2` .Let the sum of `n` terms of the series is `2046`. Thus,        `S_n=\frac{a\left(r^n-1\right)}{r-1}=\frac{2\left(2^n-1\right)}{2-1}=2^{n+1}-2=2046\implies2^{n+1}=2046+2=2048=2^11`
    `\implies`           `n+1=11\impliesn=10`
    Thus, the sum of the first `10` terms of the series is `2046`.
  3.  Find the sum of the first `n` terms of the series `4+44+444+...`
    Solution: Here, `S_n=4+44+444+....`to `n` terms `=4{1+11+111+1111+...` to `n` terms}
    `=\frac{4}{9}{9+99+999+9999+....` to `n` terms}
    `=\frac{4}{9}{\left(10-1\right)+\left(100-1\right)+\left(1000-1\right)+....` to `n` terms}
    `=\frac{4}{9}{10+100+1000+....` to `n` } `-` `{1+1+1+...` to `n` terms}
    `=\frac{4}{9}{\frac{10\left(10^n-1\right)}{10-1}-n\.1}`
    `=\frac{40}{81}\left(10^n-1\right)-\frac{4n}{9}`.
  4. Evaluate `\sum_{k=1}^{11}\left(2+3^k\right)`
    Solution: `\sum_{k=1}^{11}\left(2+3^k\right)=\sum_{k-1}^{11}2+\sum_{k-1}^{11}3^k=2\times11+\left(3+3^2+3^3+....+3^11\right)`
    `=22+\frac{3}{2}\left(3^11-1\right)`
  5. If `f` is a function satisfying `f\left(x+y\right)=f\left(x\right)f\left(y\right)`   `\forallx,y\inN` such that `f\left(1)=3` and `\sum_{x=1}^{n}f\left(x\right)=120` find the value of `n`.  
    Solution: Given that,
    `f\left(x)=f\left(x\right)f\left(y\right)\forallx,y\inN`
    `\therefore`         `f\left(1+1+1+...+1\right)` [there are `x` numbers of `1` in the bracket]
             `=f\left(1\right)f\left(1\right)f\left(1\right)....f\left(1\right)={f\left(1\right)}^x\forallx\inN` 
    `\implies`      `f\left(x\right)=3^x\forallx\inN`
    Also, given that,    `\sum_{x=1}^{n}f\left(x\right)=120\implies\sum_{x=1}^{n}3^x=120=3+3^2+3^3+...+3^n=120`
    `\implies`    `3\left(\frac{3^n-1}{3-1}\right)=120\implies3^n-1=80\implies3^n=81\implies3^n=3^4\impliesn=4`.
  6. The sum of some terms of G.P. is `315` whose first term and the common ratio are `5` and `2`, respectively. Find the last term and the number of terms.
    Solution: Let the number of terms of the given G.P. be `n`
    `\therefore`      `\frac{a\left(r^n-1\right)}{r-1}=315\implies\frac{5\left(2^n-1\right)}{2-1}=315` [Given, `a=5` and `r=2`] 
    `\implies`       `2^n-1=63\implies2^n=64=2^6\impliesn=6`
    `\therefore`      the last term `t_6=ar^5=5\.2^5==160`.
    Thus, the last term and number of terms of the G.P. is `160` and `6` respectively.
  7. If `S` be the sum, `P` the product and `R` the sum of the reciprocals of `n`terms of a G.P., prove that,   `P^2=\left(\frac{S}{R}\right)^n`. 
    Solution: Let the first term and common ratio of the G.P. be `a` and `r` respectively.
    `\thereforeS=a+ar+ar^2+.....`up to `n` terms `=\frac{a\left(r^n-1\right)}{r-1}`.
    `R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^2}+....` up to `n` terms `=\frac{1}{a}\left(\frac{1-\frac{1}{r^n}}{1-\frac{1}{r}}\right)=\frac{1}{a}\.\frac{r^n-1}{r^n}\.\frac{r}{r-1}=\frac{r^n-1}{ar^{n-1}\left(r-1\right)}`.
    `\therefore`    `\frac{S}{R}=\frac{a\left(r^n-1\right)}{r-1}\times\frac{ar^{n-1}\left(r-1\right)}{r^n-1}=a^2\.r^{n-1}`. 
    Now, `P=a\.ar\.ar^2\....ar^{n-1}=a^nr^{1+2+3+...+\left(n-1)}=a^nr^{\frac{n\left(n-1\right)}{2}}` 
    ` \thereforeP^2= \left(a^nr^{\frac{n\left(n-1\right)}{2}}\right)^2=\left(a^2r^{n-1}\right)^n=\left(\frac{S}{R}\right)^n`. [`\because\frac{S}{R}=a^2r^{n-1}`]  (Proved).  
  8. Show that th ratio of the sum of the first `n` terms of a G.P. to the sum of terms from `\left(n+1\right)`th to `\left(2n\right)`th term is `\frac{1}{r^n}`. 
    Solution: Let `a` be the first term and `r` the common ratio.
    `\therefore` the G.P. is, `a,ar,ar^2,...,ar^{n-1},ar^n,ar^{n+1},ar^{n+2},...,ar^{2n-1}`.
    Let `S_1` be the sum of the first `n` terms and `S_2` the sum of terms from `\left(n+1\right)`th to `\left(2n\right)`th terms. 
    `\therefore`       `S_1=\frac{a\left(r^n-1\right)}{r-1}`  and  `S_2=\frac{ar^n\left(r^n-1\right)}{r-1}\implies\frac{S_1}{S_2}=\frac{a\left(r^n-1\right)}{r-1}\times\frac{r-1}{ar^n\left(r^n-1\right)`
    Hence     `\frac{S_1}{S_2}=\frac{1}{r^n}`.   (Proved)

 Exercise-2(B) 

(MCQ)
A. Chose thee correct Answer

  1. How many terms are there in the series `4+8+16+...+256`?
    (a) `6`
    (b) `7`
    (c) `8`
    (d) `9`
  2. The sum of first `n` terms of a series is `3^na+b` when `a,b` are constants. Then the terms of the series are in
    (a) A.P.  
    (b) G.P.
    (c) A.P. from second terms onwards
    (d) G.P. from second terms onwards 
  3. Let `S_n` denotes the sum of first `n` terms of a G.P. If `S_{2n}=4S_n`, the ratio `\frac{S_{3n}}{S_n}`is
    (a) `13`
    (b) `7`
    (c) `12`
    (d) `9` 
  4. Sum of first `n` terms of the series `\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+...,` is
    (a) `2^n-1`
    (b) `1-2^{-n}`
    (c) `2^{-n}-n+1`
    (d) `2^{-n}+n-1`
  5. In a G.P. of even numbers of terms, the sum of all terms is `5` times the sum of the even terms. The common ratio of the G.P. is 
    (a) `4`
    (b) `\frac{1}{4}`
    (c) `2`
    (d) none of these
  6. The least value of `n`, for which the sum of the series `1+3+3^2+...+3^n` is greater than `345`, is 
    (a) `4`
    (b) `5`
    (c) `6`
    (d) `7`
B. Answer the following questions

  1. Find the sum to indicated number of terms in each of the following geometric progressions.
    (a) `0.15,0.015,0.0015,...20`terms                    [Ans: `\frac{1}{6}[1-\left(0.1\right)^20]`]
    (b) `\frac{1}{\sqrt3}+1+\sqrt3+3+...12` terms               [Ans: `364\left(1+\frac{1}{\sqrt3}\right)`]     
    (c) `256+128+64+....10` terms                     [Ans: `511\frac{1}{2}`]
  2. Find the sum of the first `n` terms and sum of the first 5 terms of the geometric series `1+\frac{2}{3}+\frac{4}{9}+...`  [Ans: `3[1-\left(\frac{2}{3}\right)^n],\frac{211}{81}`]
  3. How many terms of the G.P. `3,\frac{3}{2},\frac{3}{4},...` are needed to give the sum `\frac{3069}{512}`  [Ans: `10`]
  4. Find the sum of the following series up to `n` terms
    `5+55+555+.....`       [Ans: `\frac{50}{81}\left(10^n-1\right)-\frac{5n}{9}`]
  5. Find the sum of the following series up to `n` terms
    `0.6+0.66+0.666+.....` [Ans: `\frac{2n}{3}-\frac{2}{27}\left(1-\frac{1}{10^n}\right)`]
  6. If the sum of the first `n` terms of a G.P. is `p` and sum of the first `2n` terms is `3p`, then show that the sum of the first `3n` terms is `7p`.
  7. If the `4`th term of a G.P. is 24 and `7`th term is 192, find the sum of its first `10` terms   [Ans: `3069`]
  8. If `S_1,S_2,S_3` be respectively the sums of the first `n,2n,3n` terms of a G.P., then prove that, `S_1\left(S_3-S_2\right)=\left(S_2-S_1\right)^2`. 
  9. If `S` be the sum, `P` the product and `R` the sum of the reciprocals of the first 5 terms of a G.P., then show that `P^2=\left(\frac{S}{R}\right)^5`. 
  10. The sum of 2n terms of series in G.P. is `2R` and the sum of the reciprocals of these terms is `R`; find the continued product of the terms. [Ans: `2^n`]


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