the first and the last terms of an AP are 7 and 49 respectively if sum

1.	the first and the last terms of an AP are 7 and 49 respectively if sum of all its terms is 400. find the common difference
Answer» ANSWER: Step-by-step explanation: The FIRST term = 7 = a The Last term = 49 = a + (N-1)d,..............(1) where n is an integer and d is the common difference Sum of all terms in this ap = 200 = $\frac{n}{2} [ 2a + ( n-1) d]$ ............(2) Then, 200 = [n/2] [ 14+ (n-1)d] .............(3) 49 - 7 = (n-1)d.......(4) put 4 in 3, 200 = [n/2] [ 14 + 42] 200 = [n/2] [ 56] 200/56 = [n/2] 200/28 = n 100/14 = n 50 / 7 = n .......................(5) put 5 in 1 49 = 7 + ([50/7]-1)d 42 = [43/7]d [42/43] 7 = d d = 294 / 43

the first and the last terms of an AP are 7 and 49 respectively if sum of all its terms is 400. find the common difference

Answer»

ANSWER:

Step-by-step explanation:

The FIRST term = 7 = a

The Last term = 49 = a + (N-1)d,..............(1) where n is an integer and d is the common difference

Sum of all terms in this ap = 200 = $\frac{n}{2} [ 2a + ( n-1) d]$ ............(2)

Then,

200 = [n/2] [ 14+ (n-1)d] .............(3)

49 - 7 = (n-1)d.......(4)

put 4 in 3,

200 = [n/2] [ 14 + 42]

200 = [n/2] [ 56]

200/56 = [n/2]

200/28 = n

100/14 = n

50 / 7 = n .......................(5)

put 5 in 1

49 = 7 + ([50/7]-1)d

42 = [43/7]d

[42/43] 7 = d

d = 294 / 43

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