Question

Class 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Please answer fast . . . . . . . . . . . . . . . . . . . . . . .
Maths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Arithmetic Progression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The 2nd term of an AP is nine times the 5th term and the sum of the first eight terms is 56. Find the first term and the common difference.

Answer 1

this is the sum of arithmetic progression. I will solve it for you now.
here,
a2=9 × a5
a+(2-1)d=9 × {a+(5-1)d}
a+d=9×(a+4d)
a+d=9a + 36d
9a-a+36d-d=0
8a+35d=0_______(eq 1)

ATQ,
S8=8/2{2a+(8-1)d}
56=4{2×a+7d}
2a+7d =14________(eq 2).


now solve it by either substitution method or the addition and multiplication method.

Answer 2

Let a be the first term and d be the common difference.

It is given that the 2nd term of an AP is nine times the 5th term.

a + d = 9(a + 4d)

a + d = 9a + 36d

8a + 35d = 0   ----- (1).


Given that the sum of first 8 terms = 56.

We know that sum of n terms of an ap sn = n/2(2a + (n - 1) * d)

                                                                56 = 8/2(2a + (8 - 1) * d)

                                                                56 = 4(2a + 7d)

                                                                2a + 7d = 14   ---------- (2)


On solving (1) & (2) * 5, we get

10a + 35d = 70 

8a + 35d = 0

----------------------

2a = 70

a = 35.


Substitute a = 35 in (1), we get

8a + 35d = 0

8(35) + 35d = 0

280 + 35d = 0

d = -280/35

d = -8.


Therefore the first term of the AP is 35 and the common difference is -8.


The series will be 35,27,19,11.......


Hope this helps!