The sum of the first 7 terms of an A.P. is 10, and that of the next 7

1.	The sum of the first 7 terms of an A.P. is 10, and that of the next 7 terms is 17. Find the progression
Answer» Assuming the first term as a and common difference as d To find : the progression So, the sum of first 7 terms is given by a + a + d + a + 2d + a + 3d…. a + 6d = 10 7a + 21d = 10…(i) In the second part it is given that sum of next seven terms is 17 a + 7d + a + 8d + a + 9d…. a + 13d = 7 7a + 70d = 7…(ii) Solving (i) and (ii) we get 10 - 21d = 7 - 70d 3 = - 49d d = \(-\frac{3}{49}\) a = \(\frac{79}{49}\) Hence, The sequence is given by a, a + d, a + 2d…. which is \(\frac{79}{49}\),\(\frac{76}{49}\),\(\frac{73}{49}\),....

The sum of the first 7 terms of an A.P. is 10, and that of the next 7 terms is 17. Find the progression

Answer»

Assuming the first term as a and common difference as d

To find : the progression

So, the sum of first 7 terms is given by

a + a + d + a + 2d + a + 3d…. a + 6d = 10

7a + 21d = 10…(i)

In the second part it is given that sum of next seven terms is 17

a + 7d + a + 8d + a + 9d…. a + 13d = 7

7a + 70d = 7…(ii)

Solving (i) and (ii) we get

10 - 21d = 7 - 70d

3 = - 49d

d = \(-\frac{3}{49}\)

a = \(\frac{79}{49}\)

Hence,

The sequence is given by a, a + d, a + 2d…. which is

\(\frac{79}{49}\),\(\frac{76}{49}\),\(\frac{73}{49}\),....