in an AP , if S5 + S7 = 167 and S10 = 235, then find the AP where Sn denotes the sum of its first n terms
Answers
Answer:
Step-by-step explanation:
Answer
We know that sum of n terms sn = n/2(2a + (n - 1) * d)
Given s5 + s7 = 167.
= 5/2(2a + (5 - 1) * d) + 7/2(2a + (7 - 1) d) = 167
= 5/2(2a + 4d) + 7/2(2a + 6d) = 167
= 5(a + 2d) + 7(a + 3d) = 167
= 5a + 10d + 7a + 21d = 167
= 12a + 31d = 167 ---------- (1)
Given that s10 = 235
10/2(2a + (10 - 1) * d) = 235
5(2a + 9d) = 235
2a + 9d = 47 ------------------- (2)
On solving (1) & (2) * 6 , we get
12a + 54d = 282
12a + 31d = 167
----------------------
23d = 115
d = 115/23
d = 5.
Substitute d = 5 in (1), we get
12a + 31d = 167
12a + 31(5) = 167
12a + 155 = 167
12a = 167 - 155
12a = 12
a = 12/12
a = 1.
Therefore the AP is a, a + d, a + 2d = 1 , 1 + 5, 1 + 5(2), 1 + 5(3).......
Therefore the sum of first n terms = 1,6,11,16.......
Hope this helps!
here is the answer......
hope this would have helped you.....
all the best for your board exams.......
