Question

Question 3 Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3 (S2– S1)

Class X1 - Maths -Sequences and Series Page 199

Answer 1

hello users.....

we have given that
the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively,

we have to show that
S3 = 3 (S2– S1)

solution :-
as shown in attachment...

⭐✡⭐ hope it helps ⭐✡⭐

Answer 2

Let a is the first term
and d is the common difference of AP
then,
S₁ = n/2{2a + (n-1)d}
S₂ = 2n/2{2a + (2n-1)d}
S₃ = 3n/2{2a + (3n-1)d} ____________(1)

now,
RHS = 3(S₂ - S₁ )
= 3[ 2n/2{2a + (2n-1)d } - n/2{2a + (n-1)d}]
= 3n/2[ 2{2a + (2n-1)d } -{2a + (n-1)d}]
= 3n/2[2a(2 -1) + d{2(2n-1)- n+1}]
= 3n/2[2a + d{4n -2 + 1 - n}]
= 3n/2[2a + d(3n -1)]
= 3n/2{2a + (3n-1)d } [ from equation (1)
= S₃ = LHS

hence, LHS = RHS