Question

If 12th term of an ap is -13 and the sum of the its first four terms is 24 what is the sum of first 10 terms

Answer 1

Answer:

0

Step-by-step explanation:

T12=a +11d

-13=a+11d

a=-13-11d                            --(i)

S4=2(2(a)+3d)

24=2(2(-13-11d)+3d)

On simplification,

d=-2

and a=9

now, S10=5[2(9)+9(-2)]

Therefore

S10=0

Answer 2

The sum of first 10 terms is 43.2

Step-by-step explanation:

Formula of nth term =$a_n=a+(n-1)d$

Substitute n = 12

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

12th term of an ap is -13

-13=a+11d

a=11d+13

Formula of Sum of n terms =$S_n=\frac{n}{2}(2a+(n-1)d$

Substitute n = 4

$S_4=\frac{4}{2}(2a+(4-1)d)\\24=2(2a+3d)\\12=2a+3d\\12=2(11d+13)+3d\\12=22d+26+3d\\12=25d+26\\-14=25d\\\frac{-14}{25}=d\\a=11d+13\\a=11(\frac{-14}{25})+13\\a=6.84\\ S_n=\frac{n}{2}(2a+(n-1)d\\S_{10}=\frac{10}{2}(2(6.84)+(10-1)(\frac{-14}{25}))\\S_{10}=43.2$

Hence The sum of first 10 terms is 43.2

#Learn more:

In an AP the 5th term is 24 and 12th term is 94,then the sum of first 20 terms is ​

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