Question

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

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

It is given that the 12th term of an AP i=-13, since $a_{n}=a+(n-1)d$, therefore

$a_{12}=a+11d$⇒$-13=a+11d$                           (1)

Also, $S_{4}= S_{1}+S_{2}+S_{3}+S_{4}$

⇒$a+(a+d)(a+2d)(a+3d)=24$

⇒$4a+6d=24$                                                             (2)

From (1)  and (2),

Multiply equation (1) by 4 and then adding to equation (2), we get

$4a+44d-4a-6d=-52-24$

$38=-76$

$d=-2$

Putting the value of d in equation (1), we get

$a+11(-2)=-13$

$a-22=-13$

$a=9$

Now, sum of first n terms=$\frac{n}{2}(2a+(n-1)d)$

=$5(18+9(-2))=0$