Question

Tn=3+4n find AP and 15th term and sum of n terms​

Answer 1

Step-by-step explanation:

Answer: 525

Step-by-step explanation:

tₙ = 3 + 4n

So t₁ = 3+4 = 7

t₁₅ = 3 + 4x15 = 63

Sum of first 15 terms, Sₙ = n/2 ( t₁ + tₙ )

So S₁₅ = 15/2 (7+63)

=15/2 x 70

= 525

Answer 2

EXPLANATION.

=> Tn = 3 + 4n

=> put n = 1 = 3 + 4 = 7

=> put n = 2 = 3 + 4(2) = 11

=> put n = 3 = 3 + 4(3) = 15

=> put n = 4 = 3 + 4(4) = 19

Therefore,

sequence = 7,11,15,19 ........

First term = a = 7

common difference = d = b - a = 11 - 7 = 4

Find = 15 th term of an Ap

Nth term of an Ap

=> An = a + ( n - 1 ) d

=> 15th term = a + 14d

=> 7 + 14 X 4

=> 7 + 56

=> 63

15th term of an Ap = 63.

Sum of n terms of an Ap

$\rm \to \: s_{n} \: = \frac{n}{2} (2a \: + \: (n \: - 1)d)$

$\rm \to \: s_{n} \: = \frac{n}{2}(2 \times 7 \: + \: (n - 1)4)$

$\rm \to \: s_{n} \: = \frac{n}{2}(14 \: + 4n \: - 4)$

$\rm \to \: s_{n} \: = \frac{n}{2}(10 + 4n)$

$\rm \to \: s_{n} \: = n(5 + 2n)$