JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Sequences and Series

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An arithmetic progression is written in the following way

\begin{tabular}{ccccccc}
& & & 2 & & & \\
& & 5 & & 8 & & \\
20 & 11 & & 14 & & 17 & \\
23 & & 26 & 26 & & 29
\end{tabular}

The sum of all the terms of the

1 0^{th}

row is ______.

Answer: 1505

Solution

Row

n

contains

n

consecutive terms of the AP

2, 5, 8, 11, \dots

(

a = 2, d = 3

), the

k

-th term being

3 k - 1

.
Row

n

starts at term index

\frac{n ( n - 1 )}{2} + 1

. For

n = 10

: start index

= 46

, end index

= 55

T_{46} = 3 (46) - 1 = 137, T_{55} = 3 (55) - 1 = 164

Sum of row 10 = \frac{10}{2} (137 + 164) = 5 (301) = 1505

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Integer answer

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