Question

If the 12th term of an A.P. is - 13 and the sum of the first four terms is 24, what is the sum of the first 10 terms?

Answer 1

Answer:

$a_{12} = - 13 \\ s_{4} = 24 \\ too \: find \: s_{10} = \: \: \: \\ solve = a + 11d = - 13 \\ a = - 13 - 11d........eq1 \\ s_{4} = \frac{n}{2} (2a + (n - 1)d) \\ 24 = \frac{4}{2} (2 \times ( - 13 - 11d) + (4 - 1)d) \\ 24 = 2( - 26 - 22d + 3d) \\ 12 = - 26 - 19d \\ 12 + 26 = - 19d \\ - 19d = 38 \\ d = - \frac{38}{19} \\ d = - 2........eq2 \\ from \: eq1... \\ a = - 13 - 11( - 2) \\ a = - 13 + 22 \\ a = 9 \\ s_{10} = \frac{10}{2} (2 \times 9 + (10 - 1)( - 2)) \\ s_{10} = 5(18 + ( - 18) \\ s_{10} = 5(0) \\ \: \: \: \: \: \: =0$

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