Q195 / 270 · NDA 2026 (Phase 1) · Mathematics
If $p^{\text{th}}$ term of an AP is $k$, then what is the sum of $p^{\text{th}}$ term, $(p+q)^{\text{th}}$ term and $(p-q)^{\text{th}}$ term of the AP ?
- A$2k$
- Correct answerB$3k$
- C$4k$
- D$5k$
Solution
Let the first term of the AP be $a$ and the common difference be $d$.
The $p^{\text{th}}$ term is given by $T_p = a + (p-1)d = k$.
The $(p+q)^{\text{th}}$ term is $T_{p+q} = a + (p+q-1)d$.
The $(p-q)^{\text{th}}$ term is $T_{p-q} = a + (p-q-1)d$.
The required sum is $T_p + T_{p+q} + T_{p-q}$.
Substituting the values, we get:
$[a + (p-1)d] + [a + (p+q-1)d] + [a + (p-q-1)d]$
$= 3a + d[(p-1) + (p+q-1) + (p-q-1)]$
$= 3a + d(3p - 3)$
$= 3[a + (p-1)d]$
Since $a + (p-1)d = k$, the sum is $3k$.
Answer: $3k$
The $p^{\text{th}}$ term is given by $T_p = a + (p-1)d = k$.
The $(p+q)^{\text{th}}$ term is $T_{p+q} = a + (p+q-1)d$.
The $(p-q)^{\text{th}}$ term is $T_{p-q} = a + (p-q-1)d$.
The required sum is $T_p + T_{p+q} + T_{p-q}$.
Substituting the values, we get:
$[a + (p-1)d] + [a + (p+q-1)d] + [a + (p-q-1)d]$
$= 3a + d[(p-1) + (p+q-1) + (p-q-1)]$
$= 3a + d(3p - 3)$
$= 3[a + (p-1)d]$
Since $a + (p-1)d = k$, the sum is $3k$.
Answer: $3k$
Hence, the correct answer is option: 'B'.