the Nth term of an arithmetic progrssion is 6+5n. find the sum of the first 20 terms?

Answer 1

you need to find the first term and the 20th term

a_1= 6+5(1)=11
a_20= 6+5(20)= 106

S= n(a1+an) / 2
S= 20(11+106) / 2
S= 1170

Answer 2

Let a be the first term, a’ be the nth term.Here, a’= 6 + 5n and sum of first 20 terms, S₂ₒ a = 6 + 5(1) = 11 …………..(as n = 1, the first term ) S₂ₒ = n/2(a + a’) = 20/2( 11 + 6 + 5(20)) = 10 (17 + 100) = 1170

Answer 3

First term is 6 + 5 x 1 = 11 = a
Sn = (n/2)(2a +(n-1)d) where a is the first term and d is the common difference.
S20 = (20/2)(22 + 19d)
d is given by 6 + 5(n +1) - (6+ 5n) = 6 + 5n + 5 - 6 - 5n = 5
S20 = (20/2)(22 + 19 x 5) = 10(22 + 95) = 1170

Answer 4

You can do it as people have said, with first and last terms, since it's an a.p., or using summation theory:

Using b£a(x) as shorthand for 'Sum of x from a to b',

20£1 (6+5n) = 20£1 (6) + 20£1 (5n)
=(20 * 6) + (5*(20£1 (n)))
=120 + (5*(20*21/2)
=120+(5*210) = 120+1050 = 1170

Hope this helps.

Answer 5

Sn=n/2(a+l)

where Sn is the sum of the
first n terms,a is the first term
and l is the last term

a=6+5=11 and l= 6+5*20=106

hence, S(20)=20/2(11+106)
=10*117=1170

therefore, the sum of the first
20 terms is 1170

i hope that this helps

Answer 6

6*20 + 5(sum of 1 - 20)
= 120 + 5(0.5*20*21)
= 120 + 5(210)
= 120 + 1050
= 1170

Answer 7

6+5(20)=106....lord i dont, i think it has been years since that stuff...hope it helps

Answer 8

What they said...