Question

The 17th term of an arithmetic sequence is 8 more than its third term. If the sum of first and 19th terms is 46, write the sequence. ​

Answer 1

Answer:

⇒a

n

=a+(n−1)d

⇒a

17

=a+16d....(1)

⇒a

10

=a+9d.....(2)

⇒a

17

−a

10

=7

⇒(a+16d)−(a+9d)=7

⇒7d=7

⇒

d=1

Answer 2

Correct question:-

The 7th term of an arithmetic sequence is 8 more than its third term. If the sum of first and 19th terms is 46, write the sequence.

Given:-

• $a_{7}$ = 8 + $a_{3}$
• a + $a_{19}$ = 46.

To find:-

• The A.P.

Solution:-

According to the 1st condition:-

$\implies$ a + 6d = 8 + a+2d

$\implies$ 6d = 8 + 2d

$\implies$ 6d-2d = 8

$\implies$ 4d = 8

$\implies$ d = $\cancel\dfrac{8}{4}$ = 2

According to the 2nd condition:-

$\implies$ a + a + 18d = 46

$\implies$ 2a + 18(2) = 46

$\implies$ 2a + 36 = 46

$\implies$ 2a = 10

$\implies$ a = $\cancel\dfrac{10}{2}$ = 5

Hence, the required A.P. is:-

5, 7, 9, 11.......