In an A.P. of $99$ terms, the sum of all the odd-numbered terms is $2550$ . Then find the sum of all the $99$ terms of the A.P.

Open in App

Solution

Given,

$S_{50} = 2550$

$= \frac{50}{2} [a + a + 98 d]$

$2550 = 25 [2 a + 98 d]$

$2 a + 98 d = \frac{2550}{25} = 102$ .........(1)

$S_{99} = \frac{99}{2} [a + a + 98 d]$

From (1)

$= \frac{99}{2} [102]$

$= 5049$

Suggest Corrections

Similar questions

99

2550

99

A . P

24

30

10.5

List I

List II

\begin{matrix} List I & List II (A) & In an A.P., the series containing 99 terms, the sum of all & (P) & 5010 the odd numbered terms is 2550 . The sum of all the 99 terms of the A.P. is (B) & f is a function for which f (1) = 1 and f (n) = n + f (n - 1) & (Q) & 5049 for each natural number n \geq 2 . The value of f (100) is (C) & Suppose f (n) = {log}_{2} (3) \cdot {log}_{3} (4) \cdot {log}_{4} (5) \dots {log}_{n - 1} (n) . & (R) & 5050 Then the sum 100 \sum k = 2 f (2^{k}) equals (D) & Concentric circles of radii 1, 2, 3, \dots, 100 cms are drawn. The & (S) & 5100 interior of the smallest circle is coloured red and the annular regions are coloured alternately green and red, so that no two adjacent regions are of the same colour. The total area of the green regions in sq. cm is k π . Then k equals (T) & 5030 \end{matrix}

Join BYJU'S Learning Program