Question

Find the number of terms in an A.P whose first term and the 6th term are 3 and 23 respectively and the sum of all terms is 406

Answer 1

Arithmetic Progression

Given: first term and sixth terms are 3 and 23 respectively and the sum of all terms is 406

To find: the number of terms in the A.P.

Solution:

• Let there be n terms in the given A.P. and the common difference is d.

• Given, first term (a1) = 6.

• Also, sixth term (a6) = 23
• or, a1 + 5d = 23
• or, 3 + 5d = 23
• or, 5d = 20
• or, d = 4

• Now the sum of the A.P. containing n terms is
• S = n/2 × {2a1 + (n - 1) d}
• = n/2 × {6 + (n - 1) × 4}
• = n/2 × {6 + 4n - 4}
• = n/2 × (2 + 4n)
• = n (2n + 1)

• By the given condition,
• n (2n + 1) = 406
• or, 2n^2 + n - 406 = 0
• or, 2n^2 + 29n - 28n - 406 = 0
• or, n (2n + 29) - 14 (2n + 29) = 0
• or, (2n + 29) (n - 14) = 0

• Since n cannot be any irrational number, the value of n is 14.

Answer: The given A.P. has 14 terms.

Answer 2

Answer:

The given AP has 14 terms.

Step-by-step explanation: