Question

if the sixth term of an arithmetic progression is 37 and sum of the first six term is 147

Answer 1

a1 is the initial term of an arithmetic progression.

The common difference of successive members is d.

The nth term of an arithmetic progression.

an = a1 + ( n - 1 ) d

In this case:

n = 6

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

a1 + 5 d = 37


The sum of the n terms of an arithmetic is:

Sn = n * ( a1 + an ) / 2

S6 = 6 * ( a1 + a6 ) / 2 = 147

3 * ( a1 + a6 ) = 147

3 * ( a1 + a1 + 5 d ) = 147

3 * ( 2 a1 + 5 d ) = 147

6 a1 + 15 d = 147

Now you must solve system:

a1 + 5 d = 37

6 a1 + 15 d = 147

Try that.

The solutions are:

a1 = 12 , d = 5

Proof:

a1 = 12

a2 = 12 + 5 = 17

a3 = 17 + 5 = 22

a4 = 22 + 5 = 27

a5 = 27 + 5 = 32

a6 = 32 + 5 = 37


S6 = a1 + a2 + a3 + a4 + a5 + a6 =

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

Answer 2

the sixth term of an arithmetic progression is 37
---> a+5d = 37
5d = 37-a

sum of the first six term is 147
---> (6/2)(2a + 5d) = 147
3(2a + 37-a) = 147
a +37 = 49
a = 12
d = 5

If you had a question, take it from there

Answer 3

SLIMPLIFY 2 TAN 60+COS30 DIVIDED BY SIN 60

Answer 4

What may be the appropriate famular for this question?