Question

find the sum of first 10 terms of the AP whose 12th term is -13 and the sum of first 4 terms is 24 ​

Answer 1

Step-by-step explanation:

ANSWER

Sum of 12 terms=354

⇒

2

12

[2a+(12−1)d]=354

⇒6[2a+11d]=354

⇒2a+11d=59−(i)

∴

Sumofoddterms

Sumofeventerms

=

27

32

∴ Sum of even terms=32x & Sum of odd terms=27x

∴32x+27x=354

⇒59x=354

⇒x=6

∴ Sum of even terms=32×6=192

⇒t2+t4+t6+t8+t10+t12+192

∴a

1

=t2=a+d−(ii)

Common difference=t4−t2=a+3d−a−d=2d

∴ Sum=192

⇒

2

n

[2a+(n−1)d]=192

⇒

2

6

[2(a+d)+5×2d]=192

⇒[2a+2d+10d]=

3

192

=64

⇒2a+12d=64−(iii)

Subtracting (i) from (iii)

d=5

The common difference=5.