GO Classes CS/DA 2025 | Weekly Quiz 1 | Fundamental Course | Question: 7

Answer 1

Let $a$ be the first term and $d$ be the common difference the $\text{AP’s}$. Then sum to their $n$ terms is given by,

$S_n=\frac{n}{2}(2 a+(n-1) d)$

$S_p=q$

$\frac{p}{2}\{2 a+(p-1) d\}=2 a p+p(p-1) d=2 q  \qquad \longrightarrow (i)$

Similarly,

$\begin{aligned}
& S_q=p \\
& \frac{q}{2}\{2 a+(q-1) d\}=p \\
& 2 a q+q(q-1) d=2 p  \qquad \longrightarrow (ii)
\end{aligned}$

Subtract equation $(ii)$ from equation $(i),$ we get

$\begin{aligned}
& 2 a(p-q)+\{p(p-1)-q(q-1)\} d=2 q-2 p\\
& 2 a+(p+q-1) d=-2 \qquad \longrightarrow (iii) 
\end{aligned}$

$\begin{aligned}
& S_{p+q}=\frac{p+q}{2}\{2 a+(p+q-1) d\} \\
& S_{p+q}=\frac{p+q}{2}(-2) \qquad [\because \text{From equation} \;(iii)] \\
& \boxed{S_{p+q}=-(p+q)}
\end{aligned}$

Reference:

• https://lh6.googleusercontent.com/Qerpe-hTn1BPh_0QyQfHDxcqXEotQF-nMVKJ0I3_CtJ8PPK-WFc82FQfTlKDtJPFK2lcMxILuyQC_NP4T8Quqg9CED-mwzfI6Ka_cMnIYDiDXhBQNoDnpOdAUzofioxN4GW5-U1b

Comments

Thanks for letting us know. We have added that in question.

---

It should be given additionally that p != q @Sachin Mittal 1

---

the reference link is not working anymore!

Answer 2

@GOCLASSES

Answer 3

Correct option is D.

Sum of n terms of an A.P(S_n)= n/2[2a+(n−1)d]

As given, Sum of p terms(S_p) = q:

So, S_p=q

⇒p/2[2a+(p−1)d]=q

⇒2ap+(p−1)pd=2q      (i)

Similarly for, Sum of q terms(Sq):

S_q=p

⇒q/2{2a+(q−1)d}=p

⇒2aq+(q−1)qd=2p      (ii)

on Subtracting eq.(ii) from eq.(i):

[2ap+(p−1)pd]−[2aq+(q−1)qd]=2q−2p

⇒2ap−2aq+[(p−1)pd-(q−1)qd]=2q−2p

Expanding the above expression:

⇒2a(p−q)+p²d−pd−q²d+qd=−2(p−q)

⇒2a(p−q)+p²d−q²d−(p−q)d=−2(p−q)

⇒2a(p−q)+(p²−q²)d−(p−q)d=−2(p−q)

⇒2a(p−q)+(p−q)(p+q)d−(p−q)d=−2(p−q)   [∵(a²−b²)=(a−b)(a+b)]

Now, cancelling (p−q) from both sides

⇒2a+(p+q)d−d=−2    

⇒2a+(p+q−1)d=−2       (iii)

Now, we find sum of (p+q) terms i.e. S_p+q

⇒(p+q)/2[2a+(p+q−1)d]

Using eq.(iii) [2a+(p+q−1)d] = -2

⇒(p+q)/2[−2]

⇒−(p+q)

Answer 4

This Solution looks like lengthy but very simple and Detailed. just look once

Let a be the Term and d be the common difference