IPMAT Indore 2025 (MCQ) - Let S_1 = \100, 105, 110, 115, ... \ and S_2 = \100, 95, 90, 85, ... \ be two series in arithmetic progression. If a_k and b_k are the k -th terms of S_1 and S_2 , respectively, then _k=1^20 a_k b_k equals __________. | PYQs + Solutions

IPMAT Indore 2025

Algebra

Progression & Series

Medium

Let $S_1 = \{100, 105, 110, 115, ... \}$ and $S_2 = \{100, 95, 90, 85, ... \}$ be two series in arithmetic progression. If $a_k$ and $b_k$ are the $k$ -th terms of $S_1$ and $S_2$ , respectively, then $\sum_{k=1}^{20} a_k b_k$ equals __________.

137275

138250

137225

135375

✅ Correct Option: 2

Let the Arithmetic Progression,

S_1 = \{100, 105, 110, 115, ...\}

\newline

- first term

(a_1) = 100

\newline

- common difference

(d_1) = 5

Let the Arithmetic Progression,

S_2 = \{100, 95, 90, 85, ...\}

\newline

- first term

(b_1) = 100

\newline

- common difference

(d_2) = -5

For an arithmetic progression with first term

a

and common difference

d

, the

k

-th term is:

T_k = a + (k-1)d

For

S_1

a_k = 100 + (k-1)(5) = \boxed{95 + 5k}

For

S_2

b_k = 100 + (k-1)(-5) = \boxed{105 - 5k}

Now let's find the product

a_k b_k

a_k b_k = (95 + 5k)(105 - 5k)

\newline

a_k b_k = 9,975 + 50k - 25k^2

Calculating the sum:

\Rightarrow \sum_{k=1}^{20} a_k b_k

\Rightarrow \sum_{k=1}^{20} (9,975 + 50k - 25k^2)

This splits into three parts:

\Rightarrow 20 \times 9,975 + 50\sum_{k=1}^{20} k - 25\sum_{k=1}^{20} k^2

Part 1:

20 \times 9,975 = 199,500

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\newline

Part 2: Using

\sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}

50\sum_{k=1}^{20} k = 50 \times \dfrac{20 \times 21}{2} = 10,500

\newline

\newline

Part 3: Using

\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}

25\sum_{k=1}^{20} k^2 = 25 \times \dfrac{20 \times 21 \times 41}{6} = 71,750

Combining all parts:

\sum_{k=1}^{20} a_k b_k = 199,500 + 10,500 - 71,750 = 138,250