Question

If the 8th term of an Ap is 31 and its 15th term is 16 more than the 11th term, find the AP.

Answer 1

n11+d=n12, n12+d=n13, n13+d=n14, n14+d=n15

hence n11+4d=n15

but n15-n11=16

therefore 4d=16 i.e. d=4

now if a is first term of the AP, then a+7d=n8

but n8=31 hence a=31–7x4=3

therefore the AP is

3, 7, 11, 15, 19, 23, 27, 31…..

Answer 2

given:- 8th term is 31 so from the formula
tn=a+(n-1)d
31=a+(8-1)d
31=a+7d........(1)
now it is given that 15th term is 16 more than 11th term so
15th term =a+(n-1)d
a+(15-1)d
a+14d..........(2)
now similarly 11th term=a+10d........(3)
but according to question
a+10d=a+14d+16
solving this
a-a+10d-14d=16
-4d=16
d=-4
substitung the value of d in equation (1)we get
31=a+7(-4)
31=a-28
31+28=a
a=59

so now we get the first term and common difference,so by applying formula we can find the ap
so ap will =59,55,51,47....nd so on
you can also verify it by putting in the formula nd you will get 15th term 16 more than 11th term


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