Question

find the sum of an AP 2, 5, 8 ,11........ 62​

Answer 1

Step-by-step explanation:

Given AP:

2, 5, 8, 11.... 62

Here,

a = 2, d = 8 - 5 = 3, an = 62

Here,

We aren't provided with the number of terms of which we have to find the sum of.

But we are given with the last term of the term, by which we are surely find out the number of terms.

By using general term,

an = a + (n-1)d

62 = 2 + (n - 1)3

62 = 2 + 3n - 3

62 = - 1 + 3n

63 = 3n

$\boxed{\bold{\pink{n = 21}}}$

Now, we are having number of terms.

So, by using sum of first n terms,

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

$S_{21} = \frac{21}{2}[2(2) + (21-1)3]$

$S_{21} = \frac{21}{2}[4 + 60]$

$S_{21} = \frac{21}{2}(64)$

$S_{21} = \frac{21}{\cancel{2}}(\cancel{64})$

$S_{21} = 21 \times 32$

$\huge{\pink{\boxed{\mathsf{S_{21} = 672}}}}$

Answer 2

Answer:Given  AP:

2, 5, 8, 11.... 62

Here,

a = 2, d = 8 - 5 = 3, an = 62

Step-by-step explanation: hope it helps