if the sum of the first seven terms of an arithmetic series is 10 and the sum of the next seven terms is 17, then the fifth term of the series is:
Answers
Answer:
1, 8/7, 9/7....
Step-by-step explanation:
Let the first term be A and Common difference be D
We know that Sum of n terms = n/2[2a + (n-1)d], where
n = number of terms, a = first term, d = common difference
Given that Sum of first 7 term = 10
∴ 7/2[2A + 6D] = 10
14A + 42D = 20 ----- ( i )
Also,
Sum of Next 7 terms = 17
Sum of next 7 terms will be Sum of 14 terms - Sum of first 7 terms
∴ [14/2(2A + 13D)] - 10 = 17 [Sum of first 7 terms given 10]
7(2A + 13D) = 27
14A + 91D = 27 ------- ( ii )
Subtracting ( i ) from ( ii )
14A + 91D - (14A + 42D) = 27 - 20
49D = 7
D = 1/7
Putting Value of D in ( i )
14A + 42D = 20
14A + 42(1/7) = 20
14A + 6 = 20
14A = 14
A = 1
∴ First Term is 1 and Common Difference is 1/7
∴ AP is 1, 1+1/7, 1+2/7... or 1, 8/7, 9/7....
Step-by-step explanation:
1, 8/7, 9/7....
Step-by-step explanation:
Let the first term be A and Common difference be D
We know that Sum of n terms = n/2[2a + (n-1)d], where
n = number of terms, a = first term, d = common difference
Given that Sum of first 7 term = 10
∴ 7/2[2A + 6D] = 10
14A + 42D = 20 ----- ( i )
Also,
Sum of Next 7 terms = 17
Sum of next 7 terms will be Sum of 14 terms - Sum of first 7 terms
∴ [14/2(2A + 13D)] - 10 = 17 [Sum of first 7 terms given 10]
7(2A + 13D) = 27
14A + 91D = 27 ------- ( ii )
Subtracting ( i ) from ( ii )
14A + 91D - (14A + 42D) = 27 - 20
49D = 7
D = 1/7
Putting Value of D in ( i )
14A + 42D = 20
14A + 42(1/7) = 20
14A + 6 = 20
14A = 14
A = 1
∴ First Term is 1 and Common Difference is 1/7