If sum of n, 2n and 3n terms of A.P. are S₁, S₂ and S₃ respectively. Show that: S₃ = 3(S₂ − S₁)

**OR**

If a, b, c and d are in G.P., show that: (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)²

Question

If sum of n, 2n and 3n terms of A.P. are S₁, S₂ and S₃ respectively. Show that: S₃ = 3(S₂ − S₁)

**OR**

If a, b, c and d are in G.P., show that: (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)²

Accepted Answer

>

> [!TLDR]
> Expanding all three sums in terms of $a$, $d$, $n$ and simplifying both $S_{3n}$ and $3(S_{2n}-S_n)$ gives the same expression.
>
> Let the AP have first term $a$, common difference $d$. Using $S_k=\dfrac{k}{2}[2a+(k-1)d]$:
>
> $S_1 = S_n = \dfrac{n}{2}[2a+(n-1)d]$
>
> $S_2 = S_{2n} = \dfrac{2n}{2}[2a+(2n-1)d] = n[2a+(2n-1)d]$
>
> $S_3 = S_{3n} = \dfrac{3n}{2}[2a+(3n-1)d]$
>
> Compute $S_2-S_1$:
>
> $S_2-S_1 = n[2a+(2n-1)d] - \dfrac{n}{2}[2a+(n-1)d]$
>
> $= 2an + n(2n-1)d - an - \dfrac{n(n-1)}{2}d$
>
> $= an + nd\left[(2n-1)-\dfrac{n-1}{2}\right] = an + nd\cdot\dfrac{4n-2-n+1}{2} = an + \dfrac{n(3n-1)d}{2}$
> …

What you need	Formula
nth term	$T_n = a + (n-1)d$
Sum of n terms	$S_n = \frac{n}{2}[2a + (n-1)d]$
Sum using last term	$S_n = \frac{n}{2}(a + l)$
Common difference	$d = T_{n+1} - T_n$

Q.If sum of n, 2n and 3n terms of A.P. are S₁, S₂ and S₃ respectively. Show that: S₃ = 3(S₂ − S₁) OR If a, b, c and d are in G.P., show that: (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)²

Concept understanding — Arithmetic Progression

Arithmetic Progression: The Pattern of Equal Steps

The Core Idea

The General Term (nth term)

Why "Arithmetic"?

Sum of the First n Terms

Quick Reference

Q.If sum of n, 2n and 3n terms of A.P. are S₁, S₂ and S₃ respectively. Show that: S₃ = 3(S₂ − S₁) OR If a, b, c and d are in G.P., show that: (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)²

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