Question

Find the sum of 1st 30 terms of an ap whose 2nd term is 2 and 7th term is 22 with explanation ​

Answer 1

Hey mate,.

Ap of second term=2

And seventh term=22

By using the formula:-

An=a((n-1)×d

Here,

a2=2

where ,

a+d=2----->1

a7=22

Where,

a+6d=22----->2

Solve equation 1&2

(2-d)+6d=22

5d=22-2

5d=20

therefore,

d=20/5

d=4

replace the value of d in eq 1

a+4=2.

a=-2

By using another formula to find sum of terms

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

For 30 terms,

n=30

Sn=30/2(2(-2)+(29)×4)

Sn=15(-4+120-4)

Sn=15×112

Sn=1680

Hope it will help you with

✌️sai

Answer 2

GIVEN :-

In an AP ,

${a}_{2} = 2 \\ \\ {a}_{7} = 22$

TO FIND:-

${s}_{30}$

SOLUTION :-

${a}_{7} - {a}_{2} \\ \\ = a + (7 - 1)d - (a + (2 - 1)d) \\ \\ = a + 6d - a - d \\ \\ \\ = 5d$

Now,

${a}_{7} - {a}_{2} = 22 - 2 \\ \\ \implies 5d = 20 \\ \\ \implies d = 4$

It can used for obtaining value of a from this equation :-

${a}_{2} = a + (n - 1)d \\ \\ \implies 2 = a + (2 - 1)4 \\ \\ \implies 2 = a + 4 \\ \\ \implies \boxed{a = - 2}$

Now we can use this obtiained data to solve the question ,

${s}_{n} = \frac{n}{2} [ 2a +( n - 1)d] \\ \\ \implies {s}_{30} = \frac{30}{2} [ 2 \times - 2 +( 30 - 1)4] \\ \\ \\ = 15( - 4 + 29 \times 4) \\ \\ = 15( - 4 + 116) \\ \\ = 15(112) \\ \\ \\ = \boxed{1680}(ans)$