Question

in an ap if the 12th term is -13and the sum of first four term is _24 what is the sum of the first 10 terms

Answer 1

a(n) =a+ (n-1)d
a(12) = a+(12-1)d
-12 =a+11d
or a=11d +12....... eq1
Now a(1)+a(2)+a(3)+a(4)= 24 (given)
a+a+d+a+2d+a+3d =24
From eq1 putting the value of a we get d=-2
Hence a= 9
Sum=10/2(2*9+(10-1)-2) =0

Answer 2

Answer:

$\frac{-1560}{19}$

Step-by-step explanation:

Given :The 12th term of an ap is -13 and sum of its first four terms is 24

To find : what is the sum of its first 10 terms?

Solution :

A.P series is a+a+d,a+2d,......

a is the first term , d is the common difference

We have given,

12th term of an AP is -13

$a_1_2=a+11d$

$-13=a+11d$ ........[1]

The formula of sum of n terms is

$s_n=\frac{n}{2}(2a+(n-1)d)$

Where n is the number of terms

The sum of its first four terms is -24

$s_4=\frac{4}{2}(2a+(4-1)d)$

$-24=4a+6d$  --2

Solving 1 and 2

Substitute the value of a from 1 in 2

$-24=4(-13-11d)+6d$

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

$-24+52= -38d$

$28= -38d$

$\frac{-28}{38}= d$

$\frac{-14}{19}= d$

Put the value of d in 1 to get value of a

$-13=a+11\times\frac{-14}{19}$

$-13=a+\frac{-154}{19}$

$-13+\frac{154}{19}=a$

$\frac{-93}{19}=a$

Put value of d in [1]

Sum of first 10 terms :

$s_10=\frac{10}{2}(2\times\frac{-93}{19}+(10-1)\times\frac{-14}{19})$

$s_10=5(2\times\frac{-93}{19}+9\times\frac{-14}{19})$

$s_10=\frac{-1560}{19}$

hence the sum of first 10 terms is $\frac{-1560}{19}$