Question

In an A.P., if the 12th term is -13 and the sum of its first four terms is 24, find the sum of its

first ten terms​

Answer 1

Given that,

12th term of an AP is -13.

$\sf a_{12} = -13 \\ \longrightarrow \sf a + 11d = - 13 -----(1)$

Sum of the first four terms is 24.

$\sf S_4 = 24 \\ \\ \longrightarrow \sf \dfrac{4}{2}\{2a + (4-1)d\} = 24 \\ \\ \longrightarrow \sf 2a + 3d = 12 --------(2)$

Multiplying equation (1) with 2,

$\longrightarrow \sf 2a + 22d = - 26 --------(3)$

Solving equations (2) and (3), we obtain :

$\longrightarrow \sf (2a + 22d) - (2a + 3d) = - 26 - 12 \\ \\ \longrightarrow \sf \: 19d = - 38 \\ \\ \longrightarrow \sf |d = - 2|$

Putting d = - 2 in (1),

$\sf \: \longrightarrow a - 22 = - 13 \\ \\ \longrightarrow \sf \: |a = 9|$

Now,

$\longrightarrow \sf \: S_{10 } = \dfrac{10}{2} \{2(9) + (10 - 1)( - 2) \} \\ \\ \longrightarrow \sf \: S_{10 } = 5(18 - 18) \\ \\ \longrightarrow \boxed{ \boxed{ \sf \: S_{10 } =0}}$

Answer 2

The answers with explanation are in the attachment's.

Please do go through them.