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Answer 1

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Answer 2

Step-by-step explanation:

Question:-

The 12th term of an AP is -13 and the sum.of its first four term is 24. Find the sum of its first 10 terms.

Solution:-

As, The nth term of an AP is given by the formula :-

$\boxed{ \bf \leadsto a_{n} = a + (n - 1)d}$

The 12th term = -13

$\sf \dashrightarrow a_{12} = - 13$

$\sf \dashrightarrow a + (12 - 1)d= - 13$

$\sf \dashrightarrow a + 11d= - 13......(i)$

And, the Sum of n terms is given by the formula:-

$\boxed{ \bf \leadsto S_{n} = \frac{n}{2} \big[ 2a + (n - 1)d\big] }$

The sum of first four terms = 24

$\sf \dashrightarrow S_{4} = 24$

$\sf \dashrightarrow \dfrac{4}{2} \big[ 2a + (4- 1)d \big] = 24$

$\sf \dashrightarrow 2(2a + 3d) = 24$

$\sf \dashrightarrow 2a + 3d= 12......(ii)$

Multiply equation (i) with 2

$\sf \dashrightarrow 2(a + 11d= - 13)$

$\sf \dashrightarrow 2a + 22d= - 26......(iii)$

Equation (iii) - Equation (i)

$\sf \dashrightarrow 2a + 22d - (2a + 3d)= - 26 - 12$

$\sf \dashrightarrow 2a + 22d - 2a - 3d = -38/tex]</p><p></p><p></p><p>[tex]\sf \dashrightarrow 19d = - 38$

$\sf \dashrightarrow d = \dfrac{- 38}{2}$

$\dashrightarrow \boxed{\sf d = - 2}$

Substituting d = -2 in equation (i)

$\sf \dashrightarrow a + 11d= - 13$

$\sf \dashrightarrow a + 11(-2) = - 13$

$\sf \dashrightarrow a - 22= - 13$

$\sf \dashrightarrow a = 22 - 13$

$\dashrightarrow \boxed{\sf a = 9}$

Sum of first 10 terms:-

$\boxed{ \bf \leadsto S_{n} = \frac{n}{2} \big[ 2a + (n - 1)d\big] }$

$\sf \dashrightarrow S_{10} = \frac{10}{2} \big[ 2(9) + (10 - 1)-2\big]$

$\sf \dashrightarrow S_{10} = 5\big( 18 + (9 \times -2)\big]$

$\sf \dashrightarrow S_{10} = 5 × 0$

$\sf \dashrightarrow S_{10} = 0$

$\underline{ \boxed{ \purple{\therefore\textsf{ \textbf{Sum \: of \: 10 \: terms = 0}}}}}$