if the sum of of 6 terms of an A.p is 36 and that of 10 terms of an A.p is 100 ,find the sum of its nth terms
Answers
Given
If the sum of of 6 terms of an A.p is 36 and that of 10 terms of an A.p is 100 .
Find out
Find the sum of nth term
Solution
- S6 = 36
As we know that
→ Sn = n/2[2a + (n - 1)d]
→ S6 = 6/2[2a + (6 - 1)d]
→ 36 = 3[2a + 5d]
→ 36/3 = 2a + 5d
→ 2a + 5d = 12 ----(i)
- S10 = 100
→ S10 = 10/2[2a + (10 - 1)d]
→ 100 = 5[2a + 9d]
→ 100/5 = 2a + 9d
→ 2a + 9d = 20 -----(ii)
Subtract both the equation
→ (2a + 5d) - (2a + 9d) = 12 - 20
→ 2a + 5d - 2a - 9d = -8
→ - 4d = - 8
→ d = 8/4 = 2
Put the value of d in Equation (i)
→ 2a + 5d = 12
→ 2a + 5 × 2 = 12
→ 2a + 10 = 12
→ 2a = 12 - 10
→ 2a = 2
→ a = 1
Now the sum of nth term
- Common Difference (d) = 2
- First term (a) = 1
→ Sn = n/2[2a + (n - 1)d]
→ Sn = n/2[2 × 1 + (n - 1)2]
→ Sn = n/2[2 + 2n - 2]
→ Sn = n/2 × 2n
→ Sn = n²
Hence, the sum of nth term is n²
S6 = 36
→ Sn = n/2[2a + (n - 1)d]
→ S6 = 6/2[2a + (6 - 1)d]
→ 36 = 3[2a + 5d]
→ 36/3 = 2a + 5d
→ 2a + 5d = 12 ----(i)
S10 = 100
→ S10 = 10/2[2a + (10 - 1)d]
→ 100 = 5[2a + 9d]
→ 100/5 = 2a + 9d
→ 2a + 9d = 20 -----(ii)
Subtract both the equation
→ (2a + 5d) - (2a + 9d) = 12 - 20
→ 2a + 5d - 2a - 9d = -8
→ - 4d = - 8
→ d = 8/4 = 2
Put the value of d in Equation (i)
→ 2a + 5d = 12
→ 2a + 5 × 2 = 12
→ 2a + 10 = 12
→ 2a = 12 - 10
→ 2a = 2
→ a = 1
Common Difference (d) = 2
First term (a) = 1
→ Sn = n/2[2a + (n - 1)d]
→ Sn = n/2[2 × 1 + (n - 1)2]
→ Sn = n/2[2 + 2n - 2]
→ Sn = n/2 × 2n
→ Sn = n²