Find the sum of 12 terms of an A.P. whose nth term is given by an = 3n + 4

Sure, I can help with that. The nth term of an arithmetic progression (A.P.) is given by the formula an = a + (n-1)d, where a is the first term and d is the common difference.

In this case, the nth term is given by an = 3n + 4.

To find the first term (a1), we substitute n = 1 into the formula:

a1 = 3(1) + 4 = 7

To find the second term (a2), we substitute n = 2 into the formula:

a2 = 3(2) + 4 = 10

The common difference (d) is the difference between the second term and the first term:

d = a2 - a1 = 10 - 7 = 3

The sum (S) of the first n terms of an A.P. is given by the formula S = n/2 [2a + (n-1)d].

Substituting the values we found into this formula, the sum of the first 12 terms is:

S = 12/2 [2(7) + (12-1)(3)] = 6 [14 + 33] = 6 * 47 = 282

So, the sum of the first 12 terms of the A.P. is 282.

I'm sorry, but you didn't provide a text to respond to. Could you please provide the text?

Sure, I can help with that. The nth term of an arithmetic progression (A.P.) is given by an = 3n + 4.

The sum of the first n terms of an A.P. is given by the formula:

S_n = n/2 * (a + l)

where:

• S_n is the sum of the first n terms,
• n is the number of terms,
• a is the first term, and
• l is the last term.

First, we need to find the first term (a) and the last term (l).

The first term a1 is found by substituting n = 1 into the nth term formula:

a1 = 3*1 + 4 = 7

The 12th term a12 is found by substituting n = 12 into the nth term formula:

a12 = 3*12 + 4 = 40

Now we can substitute n = 12, a = 7, and l = 40 into the sum formula:

S_12 = 12/2 * (7 + 40) = 6 * 47 = 282

So, the sum of the first 12 terms of the A.P. is 282.