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
if the sixth term of an arithmetic progression is 37 and sum of the first six term is 147
4 answers
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
---> 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
SLIMPLIFY 2 TAN 60+COS30 DIVIDED BY SIN 60
What may be the appropriate famular for this question?