Question

​The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find its common difference.

Answer 1

Answer:

The number of terms is  15 and common difference is  3

Step-by-step explanation:

Given :  

First term, a = 7, nth term, an (l) = 49 and Sn = 420

By using the formula ,Sum of nth terms , Sn = n/2[a + l]

420 = n/2 [7 + 49]

420 = n/2 × 56

420 = 28n  

n = 420/28

n = 15

Number of terms, n = 15

By using the formula ,an = a + (n - 1)d

49 = 7 + (15 – 1) d

49 = 7 + 14d

49 - 7  = 14d

42 = 14 d

d = 42/14

d = 3

Common Difference ,d = 3  

Hence, the number of terms is  15 and common difference is  3

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Answer 2

GIVEN :

First term of an AP = a = 7

Last term of an AP = an = 49

Sum of these term = 420

Common Difference = ?

In an AP nth term = an = a + ( n - 1 )d

49 = 7 + (n - 1)d

49 = 7 + ( n - 1 )d

49 - 7 = (n - 1)d

42 = dn - d ------(1)

In an AP sum of the terms = n/2 ( a + an)

420 = n/2 ( 7 + 49 )

420 = n/2 ( 56 )

420 × 2 = 56n

840 = 56n

n = 840/56

n = 15

Number of terms = 15

Substitute n in eq - (1) to find d

42 = dn - d

42 = d(15) - d

42 = d( 15 - 1)

42 = d(14)

42 = 14d

d = 42/14

d = 3

Therefore, the common difference (d) = 3.