Question

if sum of 13 term is 169 and that of 15 term is 225 then find sum of n term.​

Answer 1

Answer:

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Answer 2

Answer:

$n^2$ is the value of sum of n term of an A.P.

Step-by-step explanation:

Explanation:

Given, sum of 13 term  = 169

and sum of 15 term = 225 .

Now, as we know that the sum of n term of an A.P ,

$S_n$ = $\frac{n}{2}[2a + (n-1 )d]$

Step 1:

Therefore, $s_{13} = \frac{13}{2}[2a + (13 -1) d]$

⇒$169 = \frac{13}{2}[2a + 12d]$

⇒ 338 = 26a + 156d

13(26) = 13 (2a + 12d)

⇒26 = 2a + 12 d ......(i)

Similarly, sum of 15 term =  225 can be written as,

$S_{15}$ = $\frac{15}{2}[2a + (15 - 1)d]$

⇒$225$ =$\frac{15}{2} [2a + 14 d]$

⇒450 = 30a + 210d

⇒30 (15) = 30 (a + 7d)

⇒ 15 = a + 7d ......(ii)

Step 2:

On Subtracting (i ) from (ii) we get,

26 = 2a + 12 d

2 (15 = a  +   7d)

(-)      = (-)    (-)

-4= - 2d ⇒ d = $\frac{-4}{-2}$ = 2 .

Now on putting d= 2 in equation (ii)we get,

15 = a + 7 (2)

⇒a = 15 - 14 = 1.

Step 3:

So, from step 2, first term a = 1 and common difference d = 2 .

Therefore sum of n term ,

$S_n$ = $\frac{n}{2}[2(1) + (n-1)(2)]$

$S_n$ = $\frac{n}{2}[2 + 2n - 2] = \frac{2n^2}{2} = n^2$

Final answer:

Hence, the value of sum of n term is $n^2$.

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