If $S_{1,} S_{2}, S_{3}$ are the sums of $n, 2 n, 3 n$ of an A.P. Show that $S_{3} = 3 (S_{2} - S_{1})$

Open in App

Solution

let the first term of the A.P. $a$ and the common difference be $d$ .
According to the question,
$\begin{matrix} S_{1} = \frac{n}{2} {2 a + (n - 1) d} . . . . . (1) S_{2} = \frac{2 n}{2} {2 a + (2 n - 1) d} . . . . . (2) S_{3} = \frac{3 n}{2} {2 a + (3 n - 1) d} . . . . . (3) \end{matrix}$
we have to prove that $S_{3} = 3 (S_{2} - S_{1})$
R.H.S= $3 (S_{2} - S_{1})$
$\begin{matrix} = 3 [\frac{2 n}{2} {2 a + (2 n - 1) d} - \frac{n}{2} {2 a + (n - 1) d}] = 3 \times \frac{n}{2} {4 a + (4 n - 2) d - 2 a - (n - 1) d} = \frac{3 n}{2} {2 a + (4 n - 2 - n + 1) d} = \frac{3 n}{2} {2 a (3 n - 1) d} = S_{3} = L . H . S S_{3} = 3 (S_{2} - S_{1}) \end{matrix}$

Suggest Corrections

Similar questions

Let sum of n, 2n, 3n terms of an A.P. be $S_{1}, S_{2}$ and $S_{3}$ respectively, show that $S_{3} = 3 (S_{2} - S_{1})$

n, 2 n, 3 n

S_{1}, S_{2}

S_{3}

S_{3} = 3 (S_{2} - S_{1})

, 2 n, 3 n

S_{1}, S_{2}, S_{3}

S_{3} = 3 (S_{2} - S_{1})

The sum of n, 2n, 3n terms of an A.P. are $S_{1}$ , $S_{2}$ , $S_{3}$ respectively. Prove that $S_{3}$ = 3( $S_{2}$ – $S_{1}$ ).

S_{1}, S_{2}, S_{3}

S_{3} = 3 (S_{2} - S_{1})

Join BYJU'S Learning Program