Question

in ap a12 = -13, s4 = 24 find s10​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

In an arithmetic progression with

First term = a

Common Difference = d

$\sf{ 1. \: \:n th \: \: term \: is = a+(n-1)d \: \: }$

$\displaystyle \: \sf{ 2.\: \: Sum \: of \: the \: first \: n \: terms \: is \: = \frac{n}{2} [ 2a + (n - 1)d] \: \: }$

GIVEN

$\sf{In \: an \: AP \: \: a_{12} = -13 \: , S_4 = 24 \: }$

TO DETERMINE

$\sf{S_{10}}$

CALCULATION

Let first term = a

Common Difference = d

$\sf{a_{12} = 12 th \: \: term = a + 11d\: }$

$\implies \: \sf{ a + 11d\: = - 13 } \: \: \: \: ......(1)$

$\sf{S_4 = Sum \: of \: the \: first \: 4 \: terms}$

$\displaystyle \: \sf{ = \frac{4}{2} \bigg[ 2a + (4 - 1)d \bigg] \: \: }$

$= \displaystyle \: \sf{2(2a + 3d) \: }$

$\implies \sf{ \: 2 \times (2a + 3d) = 24 \: }$

$\implies \sf{ \: (2a + 3d) = 12 \: }........(2)$

Now 2 × Equation (1) - Equation (2) gives

$\sf{ \: 19d = - 38\: }$

$\implies \: \sf{ \: d = - 2\: }$

From Equation (1)

$\sf{ \: a = - 13 - 11d \: }$

$\implies \: \sf{ \: a = - 13 + 22 = 9 \: }$

Hence

$\sf{S_{10} = Sum \: of \: the \: first \: 1 \: terms}$

$\displaystyle \: \sf{ = \frac{10}{2} \bigg[ 2a + (10- 1)d \bigg] \: \: }$

$= \sf{ 5 \times (2a + 9d)\: \: }$

$\displaystyle \: \sf{ = 5 \bigg[ (2 \times 9) + \{9 \times ( - 2) \} \bigg] \: \: }$

$\displaystyle \: \sf{ = 5 \times (18 - 18) }$

$\displaystyle \: \sf{ = 0 \: }$

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