Question

find the sum of first 50 terms of an A.P whose nth term is ( 3n + 7 )​

Answer 1

Here ,

nth term = 3n+7

Let n = 1

Therefore

1st term = 3×1+7

= 10

again let n= 2

2nd term = 3×2+7

= 13

Now we know that

common difference = T2 - T1

= 13-10

d = 3

and 1st term i.e a= 10

Therefore

Sum of 50term = n/2 {2a+(n-1)d}

= 10/2 {2×10 + (50-1)3}

50term = 5(20+147)

= 5(167)

= 835

Answer : 835

Answer 2

Answer:

$let \\ \: n = 1 \\ a1 = 3(1) + 7 = 3 + 7 = 10 \: (1) \\ again \: n = 2 \\ so \\ a2 = 3(2) + 7 \\ = 6 + 7 \\ = 13 \: (2) \\ again \: we \: have \\ n = 50 \\ therefore \\ a50 = 3(50) + 7 \\ = 150 + 7 \\ = 157 \: (3) \\ fron \: (1) \: and \: (2) \\ d = a2 - a1 \\ d = 13 - 10 = 3 \: (4) \\ hence \\ s50 = n \div 2(a1 + a50) \\ = 50 \div 2(10 + 157) \\ = 25 \times (167) \\ = 4175 \\ \\ your \: answer \: is \: 4175$

As we know,

AP = a, se, a3, an

AP= 10, 13, 157

so here,

n=1, 2, 3, 50

a1= 10, a2= 13

d= 13-10 = 3

Also=

Sn= n/2{2a+(n-1)d}

= 50/2 {2×10 + (50-1) × 3}

= 25 {20 + 49 ×3}

= 25 × {20 + 147}

=25 × 167

S (50)= 4175