Question

If the sum of the first p terms in the arithmetic series is equal to the sum of the first and the terms, then their first
(p + q) Show that the sum of the terms is zero. (p is not equal q)​

Answer 1

${\large{\underbrace{\underline{\bf{Correct\:Question}}}}}$

Q)) If the sum of the first p terms in an Arithmetic Series is equal to the sum of the first q terms, then show that the sum of its first (p+q) terms is zero.(p ≠ q).

${\large{\underbrace{\underline{\bf{Answer\checkmark}}}}}$

${\underline{\underline{\frak{\green\dag\:Given:}}}}$

• Given series is Arithmetic Series/Progression.
• Sum of first p terms = Sum of first q terms , $\sf S_p=S_q$

${\underline{\underline{\frak{\green\dag\:To \:Prove:}}}}$

• Sum of it's first (p+q) terms = 0 , $\sf S_{\:(p+q) } =0$

${\underline{\underline{\frak{\green\dag\:Required\:Solution:}}}}$

Let ,

• The first term of this A.P. = a

and,

• Common difference = d

We know,

$\boxed{ \sf \: sum \: of \: first \: n \: terms \: (s _{ \: n} )= \frac{n}{2} \{2a + (n - 1)d \}}$

Now ,

${\textit{\textbf{According~to~the~Question}}}$

$\rightarrow \sf \: sum \: of \: first \: p \: terms = sum \: of \: first \: q \: terms \\ \\ \rightarrow \sf \: s _{ \: p} = s _{ \: q} \\ \\ \rightarrow \: \sf\frac{p}{ \cancel2} \{2a + (p - 1)d \} = \frac{q}{ \cancel2} \{2a + (q - 1)d \} \\ \\ \rightarrow \sf \: p(2a + pd - d) = q(2a + qd - d) \\ \\ \rightarrow \sf \: 2ap + {p}^{2} d - pd = 2aq + {q}^{2} d - qd \\ \\ \rightarrow \sf\: 2ap - 2aq + {p}^{2} d - {q}^{2} d - pd + qd = 0 \\ \\ \sf \rightarrow \: 2a(p - q) + d( {p}^{2} - {q}^{2} ) - d(p - q) = 0 \\ \\ \rightarrow \sf \: 2a(p - q) + d(p - q)(p + q) - d(p - q) = 0 \\ \\ \sf \rightarrow \: (p - q) \{2a + d(p + q) - d \} = 0 \\ \\ \rm \: so$

$\rm \: either \: \\ \\ \mapsto \rm \: (p - q) = 0 \\ \\ \mapsto \rm\: p = q \: \: \\ \\ \rm \: but \: in \: question \: its \: given \: p \ne0$

$\rm \: or \: \\ \\ \mapsto \rm \: \{2a + d(p + q) - d \} = 0 \\ \\ \mapsto \rm \: \{ \: 2a + d(p + q - 1) \} = 0 \: \blue{....(i)}$

Now ,

$\sf \colon \implies \: sum \: of \: first \: (p + q )\: terms = s _{ \: (p + q)} \\ \\ \colon \sf \implies \: s _{ \: (p + q)} = \frac{(p + q)}{2} \purple{\underline{ \{2a + (p + q - 1) d \}}} \\ \\ \tt \: from \: eq(i) \\ \\ \sf \colon \implies \: s _{ \: (p + q)} = \frac{(p + q)}{2} \times \green0 \\ \\ \star\longrightarrow \: \underline{ \boxed{ \tt \pink{s _{ \: (p + q)} = \red{\bf{0}}}}} \: \: \therefore \: \sf \: proved \: .$

So,

The Sum of first (p+q) terms would be equal to zero (0) .

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