Question

If the sum of the AP 3, 7, 11, --- is 210, then the number of terms in this AP is
The sum of first in​

Answer 1

Answer:

We know that nth term tn = sum of of the first n terms - sum of the first n-1 terms= 210–171 = 39.

But tn = a+ (n-1)d = 39 => 3+(n-1)d = 39

=> (n-1)d = 39–3= 36. ………(1)

Now And = (n/2){2a+(n-1)d} = 210

=> (n/2){6+36} = 210 (using (1))

=> (n/2)(42)= 210 => n/2 = 210/42=5 => n= 2×5=10

Using n=10 in(1) we get (10–1)d = 36

=> d = 36/9= 4

So the AP is 3,7,11,15,……

mark as brainliest..

Answer 2

Step-by-step explanation:

given AP 3,7,11,15,19.....

Sum of terms = 210

Let us say number of terms is n.

Hence sum= n(2a+(n-1)d)/2

Now, first term = a = 3

d= common difference = 7-3 = 4

putting these values we get

210=n(6+4(n-1))/2

210 = n(3+2n-2)

210= n(2n +1)

2n2 + n -210 = 0

2n2 + 21n-20n-210 = 0

n(2n+21)-10(2n-21) = 0

(n-10)(2n+21)=0

So, n = 10 or n =-21/2, it can not be n because n is number of terms that can not be negative.

So correct value is n=10

2) Sum of first n terms of AP, Sn = 3n2 + 2n

Now choose n =1 and put in the above formula since

this means sum of first 1 term that is the first term only = 3+2 = 5

Now put n=2 to get the sum of first two terms = 3×4 + 2×2 = 12+4 = 16

This means

first term + second term = 16

but first term =5 as calculated above

so, second term = 16-5 = 11

So common difference becomes, 11-5 = 6

So the AP becomes, 5, 11, 17, 23......