Question

The sum of the series 31+33+.... +53 is
(A) 729
(B) 341
(C) 504
(D) 604​

Answer 1

Answer:

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Answer 2

The sum of the series is 504 (Option c)

Given: A series 31 + 33 + ... + 53

To find: Sum of the given series

Solution:

Let's consider $a$ be the first term of the series, $a_{2}$ be the second term, $d$ be the common difference, $a_{n}$ be the last term and total number of terms in the AP be $n$.

According to the given question,

$a = 31, a_{2} = 33 \ and \ a_{n} = 53$

∵ $d = a_{2} - a$

⇒ $d = 33 - 31 = 2$

We know that any general term in the AP can be expressed as:

∵ $a_{n} = a + (n-1)d$

We have considered the last term 53 as that general term.

∴ $53 = 31 + (n-1)2$

⇒ $(n-1)2 = 53 - 31 = 22$

⇒ $n - 1 = \frac{22}{2} = 11$

⇒ $n = 11 + 1 = 12$

It means that $a_{12}$ is 53 i.e., 53 is the last and 12th term of this AP. So, the number of terms in the AP ($n$) is 12.

Now, the sum of numbers in an AP can be calculated as:

∵ $S_{n} = \frac{n}{2} \ [2a + (n-1)d]$

⇒ $S_{n} = \frac{12}{2} \ [2*31 + (12-1)2]$

⇒ $S_{n} = 6 \ [62 + 22] = 6 \ [84]$

⇒ $S_{n} = 504$

Hence, the sum of the given series is 504.

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