Question

The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms.

Answer 1

In the given problem, let us take the first term as a and the common difference as d.

Here, we are given that,

`a_3 = 7` ....(1)

`a_7 = 3a_3 + 2` ......(2)

So using (1) in (2) we get

`a_7 = 3(7) + 2 `

= 21 + 2

= 23 .....(3)

Also we know

`a_n = a + (n - 1)d`

For the 3th term (n = 3)

`a_3 = a + (3 + 1)d`

7 = a + 2d   (Using 1)

a = 7 - 2d ...(4)

Similarly, for the 7^th term (n = 7),

`a_7 = a + (7 - 1)d`

24 = a + 6d       (Using 3)

a = 24 - 6d ....(5)

Subtracting (4) from (5), we get,

a - a = (23 - 6d) - (7 - 2d)

0 = 23 - 6d - 7 + 2d

0 = 16 - 4d

4d = 16

d = 4

Now, to find a, we substitute the value of d in (4),

a = 7 - 2(4)

a= 7 - 8

a= -1

So, for the given A.P  d = 4 and a = -1

So, to find the sum of first 20 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,

`S_n = n/2 [2a + (n - 1)d]`

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 20, we get,

`S_20 = 20/2 [2(-1) + (20 - 1)(4)]`

`= (10)[-2 + (19)(4)]`

= (10)[-2 + (19)(4)]

= (10)[-2 + 76]

=  (10) [74]

= 740

Therefore, the sum of first 20 terms for the given A.P. is `S_20 = 740`