Question

What is the sum of the first 10 terms of an A.P. 41, 36, 31, 26, ...?
(A) 635 (B) 600
(C) 200 (D) 185
ОА
Ов
O
D​

Answer 1

Answer:

{D} Is the correct option

Step-by-step explanation:

Ans} a=41,

common difference {d}= a2-a1=36-41=-5

n=10th

Sn=n/2{2a+{n-1}d}

= 10/2{2 x 41+{10-1}-5}

=5{82-45}

=5{37}

=185.

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Answer 2

SOLUTION

CORRECT QUESTION

The sum of the first 10 terms of an A.P.

41, 36, 31, 26, ...

(A) 635

(B) 600

(C) 200

(D) 185

FORMULA TO BE IMPLEMENTED

Sum of first n terms of an arithmetic progression

$\displaystyle \sf = \frac{n}{2} \bigg[2a + (n - 1)d \bigg]$

Where First term = a

Common Difference = d

EVALUATION

Here the given arithmetic progression is

41 , 36 , 31 , 26 , . . . .

First term = a = 41

Common Difference = d = 36 - 41 = - 5

Number of terms = n = 10

Hence the required sum

$\displaystyle \sf = \frac{n}{2} \bigg[2a + (n - 1)d \bigg]$

$\displaystyle \sf = \frac{10}{2} \bigg[(2 \times 41) - 5 \times (10 - 1) \bigg]$

$\displaystyle \sf = 5 \times \bigg[82 - 45 \bigg]$

$\displaystyle \sf = 5 \times 37$

$\displaystyle \sf = 185$

FINAL ANSWER

Hence the correct option is (D) 185

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