Question

4. How many terms of the AP 9.17.25.... must be taken to give a sum of 636​

Answer 1

Mate it is given that,

a=9

d=8

Sn=636

so by using formula we get,

Sn=n/2(2a+(n-1)d)

636=n/2(18+8n-8)

636*2=n(10+8n)

1272=10n+8n^2

8n^2+10n-1272=0

4n^2+5n-636=0

4n^2+53n-48n-636=0

n(4n+53)-12(4n+53)=0

(n-12)(4n+53)=0

n=12 & n=-53/4

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Answer 2

Solution :-

As provided

A.P

= 9 , 17 , 25 , .......

Sₙ = 636

So

Common difference

= 17 - 9

= 8

Now as we know that

Sum of n terms of AP

$\sf{ = \dfrac{n}{2} \times \left( 2a + (n-1)d \right) }$

So

$\rightarrow \sf{ 636 = \dfrac{n}{2} \times \left( 2(9) + (n-1)(8) \right) }$

$\rightarrow \sf{636 = \dfrac{n}{2} \times \left( 18 + 8n - 8 \right) }$

$\rightarrow \sf{ 636 = \dfrac{n}{2} \times ( 8n + 10)}$

$\rightarrow \sf{636 = 4n^2 + 5n }$

$\rightarrow \sf{ 4n^2 + 5n - 636 = 0}$

$\rightarrow \sf{ 4n^2 - 48n + 53n - 636 = 0}$

$\rightarrow \sf{4n(n - 12) + 53(n - 12) = 0}$

$\rightarrow \sf{(4n + 53)(n-12) = 0}$

So n = -(53)/4 or 12

Now as number of terms can not be negative

Hence n = 12