Question

Given n=12, l=71, d=3 in an A.P. , find a and s.​

Answer 1

Answer:

Tñ = l-(n-1)× d

Tñ=71-(11)×3

Tñ= 71-33

Tñ= 38

Tñ= a+ (n-1) d

38=a+33

a= 5

Sñ= n/2(a+l)

Sñ= 11/2(5+71)

Sñ= 5.5×76

Sñ= 418

Thus, a=5 and Sñ= 418. Ans.

Answer 2

Given that : n = 12, l = 71, d = 3

As we know that : l = a + (n-1)d

=> 71 = a + (12-1)3

=> 71 = a + 33

=> a = 71 - 33 = 38

Again, we know that :

$s_{n} = \frac{n}{2} (2a + (n - 1)d) \\ \\ = > s_{n} = \frac{12}{2} (2 \times 38 + (12 - 1)3) \\ \\ = > s_{n} = 6(76 + 33) = 654$

So, the sum of the A.P. will be 654

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