Question

Sum of first 7 terms of AP is 49 and that of next 10 terms is 240 , find the sum of first n terms.​

Answer 1

GIVEN :-

• Sum of first 7 terms of A.P is 49.

• Sum of next 10 terms of the A.P is 240.

TO FIND :-

• Sum of first 'n' terms.

FORMULA USED :-

$\\ \boxed{ \bf \: s_{(n) = } \dfrac{n}{2} \{2a + (n - 1)d \} }$

Here ,

• s(n) = Sum of 'n' terms.
• n = Number of terms.
• a = 1st term.
• d = Common difference.

SOLUTION :-

♦ Sum of first 7 terms is 49.

Here ,

• s(n) = 49
• n = 7

Putting values in formula , we get...

$\\ \sf \: 49 = \dfrac{7}{2} \{2a + (7 - 1)d \} \\ \\ \\ \sf \: \dfrac{49 \times 2}{7} = 2a + 6d \\ \\ \\ \sf \: 14 = 2a + 6d \: \: \: \: \: \: \: \: \: - - - (i)$

Also , sum of next 10 terms is 240.

Hence ,

→ Sum of first 17 terms = sum of 7 terms + sum of next 10 terms

→ Sum of first 17 terms = 49 + 240

→ Sum of first 17 terms = 289

Here ,

• s(17) = 289
• n = 17

Putting values we get...

$\\ \sf \: 289 = \dfrac{17}{2} \{2a + (17 - 1)d \} \\ \\ \\ \sf \: \dfrac{289 \times 2}{17} = 2a + 16d \\ \\ \\ \sf \: 34 = 2a + 16d \: \: \: \: \: \: \: \: \: \: \: \: \: - - - (ii)$

Subtracting equation (i) by equation (ii) , we get..

$\\ \sf \: 34 - 14 = 2a + 16d - (2a + 6d) \\ \\ \sf \: 20 = 2a + 16d - 2a - 6d \\ \\ \sf \: 20 = 10d \\ \\ \boxed { \sf \:d = 2 }$

Putting value of d = 2 in equation (i) , we get...

$\\ \sf \: 14 = 2a + 6(2) \\ \\ \sf \: 14 = 2a + 12 \\ \\ \sf \: 14 - 12 = 2a \\ \\ \sf \: 2 = 2a \\ \\ \boxed{ \sf \:a = 1 }$

Now , we will find sum of first 'n' terms.

$\\ \implies \sf \: s_{(n)} = \dfrac{n}{2} \{2a + (n - 1)d \}$

Substituting values of 'a' and 'd' we get..

$\\ \implies \sf \: s_{(n)} = \dfrac{n}{2} \{2(1) + (n - 1)2 \} \\ \\ \\ \implies\sf \: s_{(n)} = \dfrac{n}{2} \{2 + 2n - 2 \} \\ \\ \\ \implies\sf \: s_{(n) } = \dfrac{n}{ \cancel2} \{ \cancel{2} n \} \\ \\ \\ \implies\boxed { \boxed{ \sf \: s_{(n)} = {n}^{2} }}$

Hence , sum of first 'n' terms is n².