Question

the 12th term of an Ap is-13 and the sum of its first four terms is 24 . find the sum of its 10th term . please give me solution step by step by (actually mai abhi ye question kar rahi hoo or mujhe bich mai thodi confusion hai so please send the right answer now) this question is from RS
can anybody give the Answer!!!​

Answer 1

Answer :-

Sum of first 10 terms of given A.P is Zero .

Given :-

12th term of AP is -13.

Sum of first 4 terms of AP is 24.

To find :-

Sum of first 10 terms of AP .

Solution :-

It's given that 12th term of AP is -13 . Is can be written as :-

$\sf{\implies a_{12} = a + 11d = - 13}$

$\sf{\implies a + 11d = -13 \:\:\:\:\: eq \: 1st}$

Now we are given sum of first 4 terms of AP.

$\sf{\implies S_4 = \frac{n}{2} [2a + ( n-1)d ] }$

$\sf{\implies 24 = \frac{4}{2} [ 2a + (4-1)d]}$

$\sf{\implies 24 = 2 [ 2a + 3d]}$

$\sf{\implies \frac{24}{2} = 2a + 3d }$

$\sf{\implies 12 = 2a + 3d\;\:\:\: eq \: 2nd }$

Multiplying 1st equation by 2 .

$\sf{\implies -13(2) = 2[ a + 11d]}$

$\sf{\implies -26 = 2a + 22d\:\:\:\: eq\:3rd}$

Now Substracting eq 2nd from eq 3rd :-

$\sf{\implies -26 - 12 = 2a + 22d - 2a - 3d}$

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

$\sf{\implies \frac{-38}{19} = d = -2 }$

• $\underline{\underline{\sf{\implies d = -2}}}$

Putting the value of d in eq 1st .

$\sf{\implies -13 = a + 11(-2) → \: a - 22 }$

$\sf{\implies -13 + 22 = a → \: 9}$

• $\underline{\underline{\sf{\implies a = 9}}}$

Now applying formula for sum of n terms of AP .

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

$\sf{\implies S_{10} = 5 [ 18 - ( 9)-2] → \: 5 [ 18 - 18]}$

$\sf{\implies S_{10} = 5[0] = 0 }$

So sum of first 10 terms of AP is zero.

Answer 2

Answer:

This is the answer

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