If the sum of p terms of an A.P. is q and the sum of q terms is p, then the sum of (p + q) terms will be
A. 0
B. p - q
C. p + q
D. -(p + q)
Answer: Option D
Solution (By Examveda Team)
$$\eqalign{ & {\text{Sum}}\,{\text{of}}\,p\,{\text{terms}} = q \cr & i.e., \cr & {S_p} = \frac{p}{2}\left[ {2a + \left( {p - 1} \right)d} \right] = q \cr & \Rightarrow p\left[ {2a + \left( {p - 1} \right)d} \right] = 2q \cr & \Rightarrow 2ap + p\left( {p - 1} \right)d = 2q\,.....\left( 1 \right) \cr & {\text{and}}\,{\text{sum}}\,{\text{of}}\,q\,{\text{terms}} = p \cr & i.e., \cr & {S_q} = \frac{q}{2}\left[ {2a + \left( {q - 1} \right)d} \right] = p \cr & \Rightarrow q\left[ {2a + \left( {q - 1} \right)d} \right] = 2p \cr & \Rightarrow 2aq + q\left( {q - 1} \right)d = 2p\,.....\left( 2 \right) \cr & {\text{Subtracting}}\,\left( 2 \right)\,{\text{from}}\,\left( 1 \right) \cr} $$$$ \Rightarrow 2a\left( {p - q} \right) + \left\{ {{p^2} - p - {q^2} + q} \right\}d$$ $$ = 2q - 2p$$
$$ \Rightarrow 2a\left( {p - q} \right) + \left\{ {{p^2} - {q^2} - \left( {p - q} \right)} \right\}d$$ $$ = - 2\left( {p - q} \right)$$
$$ \Rightarrow 2a\left( {p - q} \right) + \left\{ {\left( {p + q} \right)\left( {p - q} \right) - \left( {p - q} \right)} \right\}d$$ $$ = - 2\left( {p - q} \right)$$
$$ \Rightarrow 2a\left( {p - q} \right) + \left( {p - q} \right)\left[ {p + q - 1} \right]d$$ $$ = - 2\left( {p - q} \right)$$ $${\text{Dividing}}\,{\text{by}}$$ $$\,\left( {p - q} \right)$$
$$\eqalign{ & \Rightarrow 2a + \left( {p + q - 1} \right)d = - 2\,.....\left( 3 \right) \cr & {S_{p + q}} = \frac{{p + q}}{2}\left[ {2a + \left( {p + q - 1} \right)d} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{p + q}}{2}\left[ { - 2} \right]\,\,\,\left[ {{\text{From}}\,\left( 3 \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = - \left( {p + q} \right) \cr} $$
This Question Belongs to Arithmetic Ability >> Progressions
Starting Point
We begin with the two equations derived from the problem:
2q = p[2a + (p-1)d]
2p = q[2a + (q-1)d]
We will now perform the operation: Equation (1) - Equation (2).
Step 1: Set up the Subtraction
We subtract the left sides from each other and the right sides from each other.
2q - 2p = p[2a + (p-1)d] - q[2a + (q-1)d]
Step 2: Simplify the Left-Hand Side (LHS)
This part is straightforward. We just factor out the common 2.
2(q - p)
Step 3: Simplify the Right-Hand Side (RHS)
This is the main calculation. We will do this one piece at a time.
A. First, expand the brackets by distributing the p and q:
p[2a + (p-1)d] becomes 2ap + p(p-1)d
q[2a + (q-1)d] becomes 2aq + q(q-1)d
So the RHS is:
(2ap + p(p-1)d) - (2aq + q(q-1)d)
Remove the parentheses, making sure to distribute the negative sign:
2ap + p(p-1)d - 2aq - q(q-1)d
B. Group the terms that have 2a together and the terms that have d together:
(2ap - 2aq) + (p(p-1)d - q(q-1)d)
C. Factor out the common parts from each group:
From the first group, factor out 2a: 2a(p - q)
From the second group, factor out d: [p(p-1) - q(q-1)]d
Now the RHS looks like this:
2a(p - q) + [p(p-1) - q(q-1)]d
D. Simplify the complex part in the square brackets [...]:
Expand the contents: [p² - p - q² + q]
Group the squared terms and the non-squared terms: [(p² - q²) - (p - q)]
Factor the "difference of squares" (p² - q²): [(p - q)(p + q) - (p - q)]
Notice that (p - q) is now a common factor. Factor it out: (p - q)[(p + q) - 1]
This simplifies to: (p - q)[p + q - 1]
E. Put the fully simplified RHS back together:
The RHS is: 2a(p - q) + (p - q)[p + q - 1]d
Step 4: Combine the Simplified Sides and Solve
Now we set our simplified LHS equal to our simplified RHS:
2(q - p) = 2a(p - q) + (p - q)[p + q - 1]d
To simplify further, we want to divide by (p - q). Let's rewrite the left side first, noting that (q - p) is the same as -1 * (p - q):
-2(p - q) = 2a(p - q) + (p - q)[p + q - 1]d
Now, we can divide every single term in the equation by the common factor (p - q):
-2 = 2a + [p + q - 1]d
This is the final result of the subtraction and simplification, which is the key expression needed to solve the problem.