First slide

Arithmetic progression

Question

$Let A be the sum of the first 20 terms and ' B' be the sum of the first 40-$ $terms of the series 1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + 5^{2} + {2.6}^{2} + . . If B - 2 A = 100 λ then λ is$

Moderate

A

248
B

464
C

496
D

232

Solution

$B-2A= \sum_{n=1}^{40} a_{n} -2 \sum_{n=1}^{20} a_{n} = \sum_{n=21}^{40} a_{n} - \sum_{n = 1}^{20} a_{n}$
$B-2A= (21^{2} + {2.22}^{2} + 23^{2} + ... + 40^{2}) - (1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + ... + 20^{2})$
$\begin{array}{l} = 20 (22 + 2.24 + 26 + 2.28 + . . + 60) \\ = 20 {\underset{20 t e r m s}{\underset{⏟}{(22 + 24 + 26 + . . + 60)}} + \underset{10 t e r m s}{\underset{⏟}{(24 + 28 + . . + 60)}}} \\ = 20 \{\frac{20}{2} (22 + 60) + \frac{10}{2} (24 + 60)\} \\ = 10 (20.82 + 10.84) \\ = 10 (1640 + 840) \\ = 100 (164 + 84) = 100 \times 248 \\ \to λ = 248 \end{array}$

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