If the sixth term of an arithmetic progression (A.P) is 37 and the sum of the six term is 147, find the first term, common difference, sum of the first fifteen terms

Answer 1

Just use the formulas you know. You have only learned a few rules so far, so the problems will have to use the few you already have.

a + 5d = 37
6/2 (2a+5d) = 147
so, a=12, d=5, and
S15 = 15/2 (2*12 + 14*5) = 705

Answer 2

In an arithmetic progression nth term:

an = a1 + ( n - 1 ) d

a1= initial term

d = common difference

a6 = a1 + ( 6 - 1 ) d = 37

a1 + 5 d = 37

Sum of n terms in A.P:

Sn = n ∙ [ 2 a1 + ( n - 1) ∙ d ] / 2

In this case n = 6

S6 = 6 ∙ [ 2 a1 + ( 6 - 1) ∙ d ] / 2 = 147

3 ∙ [ 2 a1 + 5 d ] = 147

6 a1 + 15 d = 147

Now you have system of two equations:

a1 + 5 d = 37

6 a1 + 15 d = 147
___________

Try to solve it.

The solutons are:

a1 = 12 , d = 5

Sum of the first fifteen terms:

Sn = n ∙ [ 2 a1 + ( n - 1) ∙ d ] / 2

S15 = 15 ∙ [ 2 ∙ 12 + ( 15 - 1) ∙ 5 ] / 2

S15 = 15 ∙ [ 24 + 14 ∙ 5 ] / 2

S15 = 15 ∙ ( 24 + 70 ) / 2

S15 = 15 ∙ 94 / 2

S15 = 1410 / 2

S15 = 705

Check of results.

Your A.P is:

12 , 17 , 22 , 27 , 32 , 37 , 42 , 47 , 52 , 57 , 62 , 67 , 72 , 77 , 82

Sum of first 6 terms in:

S6 = 12 + 17 + 22 + 27 + 32 + 37 = 147

S15 = 12 + 17 + 22 + 27 + 32 + 37 + 42 + 47 + 52 + 57 + 62 + 67 + 72 + 77 + 82 = 705