Question

how many terms of the AP: 9, 17, 25.........must be taken to give a sum of 636?​

Answer 1

GivEn:-

• AP :- 9, 17, 25,.....

To find:-

• Number of term must be taken to give a sum of 636?

SoluTion:-

Lets the number of terms required to give the sum of 636 be n and the common difference b/w each terms be d.

GivEn AP (Arithmetic Progression) = 9, 17, 15,...

Therefore,

• First term of AP (a) = 9

• Common difference (d) = $\sf a_2 - a$ = 17 - 9 = 8

As we know that,

$\begin{lgathered}\boxed{\begin{minipage}{20 em} \sf \displaystyle \bullet a_n=a + (n-1)d \\\\\\ \bullet S_n= \dfrac{n}{2} \left(a + a_n\right) \end{minipage}}\end{lgathered}$

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Therefore,

☯ Sum of n terms are,

$\sf S_n = \dfrac{n}{2} \bigg( a + a_n \bigg) \\\\\\ :\implies\sf S_n = \dfrac{n}{2} \bigg( 2a + (n - 1)d \bigg)\qquad\lgroup\bigg\bf a_n = a + (n - 1)d \bigg\rgroup \\\\\\ \small\sf\;\;\star\; \underline{Here,\;sum\;of\;given\;AP\;is\;636.} \\\\\\ :\implies\sf 636 = \dfrac{n}{2} \bigg(2 \times 9 + (n - 1)8 \bigg) \\\\\\ :\implies\sf 636 \times 2 = n(18 + 8n - 8) \\\\\\ :\implies\sf 2 = n(8n + 10) \\\\\\ :\implies\sf 1272 = 8n^2 + 10n \\\\\\ :\implies\sf 1272 = 2(4n^2 + 5n) \\\\\\ :\implies\sf \cancel{ \dfrac{1272}{2}} = 4n^2 + 5n \\\\\\ :\implies\sf 636 = 4n^2 + 5n$

$:\implies\sf 4n^2 + 5n - 636 = 0 \\\\\\ :\implies\sf 4n^2 + 53n - 48n - 636 = 0 \\\\\\ :\implies\sf 4n^2 - 48n + 53n - 636 = 0 \\\\\\ :\implies\sf 4n(n - 12) + 53(n - 12) = 0 \\\\\\ :\implies\sf (n - 12)(4n + 53) = 0 \\\\\\ \small\sf\;\;\star\; \underline{Both\;(n - 12)\;and\;(4n + 53)\;are\;equals\;to\;0} \\\\\\ :\implies\sf (n - 12) = 0 \;;\; (4n + 53) = 0 \\\\\\ :\implies\sf n = -12 \;;\; \dfrac{-53}{4}$

✠ n must be a natural numbers ( n ≠ -ve integer).

Hence,

$\dag$ Number of terms required to give the sum of 636 is 12.

$\dag\;{\underline{\underline{\bf{Hence\;Solved!!}}}}$

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Answer 2

Answer:

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