Question

The ratio of the 10^th term to its 30^th term of an A.P. is 1 : 3 and the sum of its first six terms is 42. Find the first term and the common difference of A.P.

Answer 1

Given,

T₁₀ : T₃₀ = 1 : 3, S₆ = 42

Let a be the first term and d be a common difference, then

\(\frac {a + 9d}{a + 29 d} = \frac 13\)

\(3a+ 27d = a + 29 d\)

\(3a -a = 29d - 27 d\)

\(2a = 2d\)

\(\therefore a = d\)

Now, S₆ = 42

\(S_ n = \frac n2 [2a + (n - 1)d]\)

\(42 = \frac 62 [2a + (6- 1)d]\)

\(42 =3 [2a +5d]\)

\(14=2a +5d\)

\(14=2a +5d\quad [\because a = d]\) 

\(14 =7a\)

\(\therefore a = \frac {14}7 = 2\)

\( \therefore a= d = 2\)

Therefore, the first term of the AP is 2 and the common difference of A.P. is 2.