Question

The 6th term of an AP is zero. Prove that the 30th term of an AP is 3 times its 14th term.
(Class 10 Maths Sample Question Paper)

Answer 1

Solution:
GIVEN:
a6 = 0

Let ‘a’ be the first term and 'd’ be the common difference on given AP.

General term of an AP:
an = a +(n -1) d
a6 = a + (6-1)d
a6 = a +5d……………(1)

a30 = a +(30-1)d
a30 = a +29d………(2)

a14 = a + (14-1)d
a14 = a +13d……………(3)
a6 = a +5d
[From eq 1]
0 = a +5d    (given)
a= -5d

Put the value of a equation to 2 & 3.
a30 = a +29d
a30 = -5d +29d
a30 = 24d
a30 = 3(8d)..................(4)

a14 = a +13d
a14 = -5d +13d
a14 = 8d………………..(5)

From eq 4 & 5
a30 = 3a14

HOPE THIS WILL HELP YOU...

Answer 2

let first term of the AP be a and the common difference be d
since 6th term is 0 then
a+(6-1)d=0
a+5d=0

30th term= a +29d
(a+5d)+24d= 0+24d=24d
also 14th term = a+13d
=(a+5d)+8d
=0+8d
=8d

3 times 14th term = 3×8d
=24d
=30th term



hence proved