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.

Accepted Answer

Let us take the first term as $a$ and the common difference as $20$ .

Using the given data, we can write,

$a_{3} = 7 \dots \dots (1)$

$a_{7} = 3 a_{3} + 2 \dots \dots (2)$

So, substituting $(1)$ in $(2)$ , we get,

$a_{7} = 3 (7) + 2$

$= 21 + 2$

$= 23 \dots \dots (3)$

Also, we know, $a_{n} = a + (n - 1) d$ .

$a_{3} = a + (3 - 1) d$

$7 = a + 2 d$ [Using $(1)$ ]

$a = 7 - 2 d \dots \dots (4)$

Similarly, for the $7^{t h}$ term, $a_{7} = a + (7 - 1) d$ .

$24 = a + 6 d$ [Using $3$ ]

$a = 24 - 6 d \dots \dots (5)$

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

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

$0 = 23 - 6 d - 7 + 2 d$

$0 = 16 - 4 d$

$4 d = 16$

$d = 4$

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., $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ .

$S_{20} = \frac{20}{2} [2 (- 1) + (20 - 1) (4)]$

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

$= (10) [- 2 + 76]$

$= (10) [74]$

$= 740$

Hence, the sum of first $20$ terms for the given A.P. is $S_{20} = 740$ .

R. D. Sharma Solutions for Chapter: Arithmetic Progression, Exercise 6: EXERCISE 5.6

R. D. Sharma Mathematics Solutions for Exercise - R. D. Sharma Solutions for Chapter: Arithmetic Progression, Exercise 6: EXERCISE 5.6

Questions from R. D. Sharma Solutions for Chapter: Arithmetic Progression, Exercise 6: EXERCISE 5.6 with Hints & Solutions

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