Question

The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is
(a) 5
(b) 10
(c) 12
(d) 14
(e) 20

Answer 1

Answer:

The number of terms in the AP are 14.

Among the given options option (d) 14  is a correct answer.

Step-by-step explanation:

Given :

A.P are 3,7, 11,15 ……. & Sn =  406

Here, a = 3 d = 7 - 3 = 4

By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]

406 = n/2  [2(3) + (n – 1) 4

406 = n/2 [6 - 4n - 4]

406  = n/2 [2 - 4n]

406  = n/2  × 2 [1 - 2n]

406  = n [1 - 2n]

406 = n - 2n²

2n² + n - 406 = 0

2n² + 29 n – 28 n – 406 = 0

[By middle term splitting]

n (2n + 29) – 14(2n + 29) = 0

(n – 14) (2n + 29) = 0

(n – 14) or (2n + 29) = 0

n = 14 or n = - 29/2 (terms can't be negative)

Hence, the number of terms in the AP are 14.

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Answer:

$\huge{\red{\boxed{\boxed{\green{\sf{14-TERMS}}}}}}$

Step-by-step explanation:

$\huge{\red{\boxed{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}}$

□A.P=3,7,11,15,...

□a1=3

□common difference=4

□Sn=406

To find:

☆number of terms in this A.P=?

$sn = \frac{n}{2} (2a + (n - 1)d) \\ = )406 = \frac{n}{2} (2 \times 3 + (n - 1) \times 4) \\ = )406 = n(3 + (n - 1) \times 2) \\ = )406 = n(3 + 2n - 2) \\ = )406 = n(2n + 1) \\ = )406 = 2 {n}^{2} + n \\ = )2 {n}^{2} + n - 406 = 0 \\ \\ use \: quadratic \: formula \\ d = {b}^{2} - 4ac \\ d = {1}^{2} - 4(2 \times - 406) \\ d = 1 - 4( - 812) \\ d = 1 + 3248 \\ d = 3249 \\ \\ x = \frac{ - b + - \sqrt{d} }{2a} \\ x = \frac{ - 1 + \sqrt{3249} }{2 \times 2} \\ x = \frac{ - 1 + 57}{4} \\ x = \frac{56}{4} \\ x = 14 \\ \\ x = \frac{ - 1 - 57}{4} \\ x = \frac{ - 58}{4} \\ value \: of \: n \: cannot \: be \: in \: fraction \: or \: in \: negative \: \\ \therefore \: \: number \: of \: terms = 14$

$\huge{\red{\boxed{\boxed{\green{\sf{14-TERMS}}}}}}$

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