If a=3 d=4 sn =406 find the value n

1.	If a=3 d=4 sn =406 find the value n
Answer» $\huge\underline\mathbb{\red S\pink{O}\purple{L} \blue{UT} \orange{I}\green{ON :}}$ ★ Given that, a = 3 d = 4 Sn = 406 ★ To find, N = ? By using sum of n terms of an AP... ☯ Sn = n/2 [ 2a + (n - 1)d ] ➡ 406 = n/2 [ 2(3) + (n - 1)4 ] ➡ 406 × 2 = n [ 6 + 4N - 4 ] ➡ 812 = n [ 2 + 4n ] ➡ 812 = 2n + 4n² ➡ 4n² + 2n - 812 = 0 ➡ 2[2n² + n - 406 ] = 0 ➡ 2n² + n - 406 = 0/2 ➡ 2n² + n - 406 = 0 ➡ 2n² + 29n - 28n - 406 = 0 ➡ n(2n + 29) - 14(2n + 29) = 0 ➡ (n - 14) (2n + 29) = 0 ➡ n - 14 = 0 ; 2n + 29 = 0 ➡ n = 0 + 14 ; 2n = 0 - 29 ➡ n = 14 ; 2n = - 29 ➡ n = 14 ; n = -29/2 Values of n can't be in fractions.So, n will be 14. ★ Verification : Substitute the value of n in the formula. ➡ S14 = 14/2 [ 2(3) + (14 - 1)(4) ] ➡ S14 = 7 [ 6 + 13(4) ] ➡ S14 = 7 [ 6 + 52 ] ➡ S14 = 7 [ 58 ] ➡ S14 = 406. Hence, it is verified... $\underline{\boxed{\bf{\purple{∴ The\;number\;of\;terms\;are\;“\;14\;”.}}}}$ Step-by-step EXPLANATION: $# PLEASE MARK ME AS BRAINLIEST✌✌✌$

Answer»

$\huge\underline\mathbb{\red S\pink{O}\purple{L} \blue{UT} \orange{I}\green{ON :}}$

★ Given that,

★ To find,

By using sum of n terms of an AP...

☯ Sn = n/2 [ 2a + (n - 1)d ]

➡ 406 = n/2 [ 2(3) + (n - 1)4 ]

➡ 406 × 2 = n [ 6 + 4N - 4 ]

➡ 812 = n [ 2 + 4n ]

➡ 812 = 2n + 4n²

➡ 4n² + 2n - 812 = 0

➡ 2[2n² + n - 406 ] = 0

➡ 2n² + n - 406 = 0/2

➡ 2n² + n - 406 = 0

➡ 2n² + 29n - 28n - 406 = 0

➡ n(2n + 29) - 14(2n + 29) = 0

➡ (n - 14) (2n + 29) = 0

➡ n - 14 = 0 ; 2n + 29 = 0

➡ n = 0 + 14 ; 2n = 0 - 29

➡ n = 14 ; 2n = - 29

➡ n = 14 ; n = -29/2

Values of n can't be in fractions.So, n will be 14.

★ Verification :

Substitute the value of n in the formula.

➡ S14 = 14/2 [ 2(3) + (14 - 1)(4) ]

➡ S14 = 7 [ 6 + 13(4) ]

➡ S14 = 7 [ 6 + 52 ]

➡ S14 = 7 [ 58 ]

➡ S14 = 406.

Hence, it is verified...

$\underline{\boxed{\bf{\purple{∴ The\;number\;of\;terms\;are\;“\;14\;”.}}}}$

Step-by-step EXPLANATION:

$# PLEASE MARK ME AS BRAINLIEST✌✌✌$

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