Question

please answer the question ​

Answer 1

Answer:

Given Sn = 2n²-15n

Then S¹ = 2(1)²-15(1)=-13

S² = 2(2)²-15(2)=-22

As we know S¹ = A¹

S²-S¹ = A²

Therefore A¹ = - 13

A²= - 22-(-13) =-9

D = A²-A¹ =-9-(-13)=+4

D=+4

A = - 13

Now put this is the formula sn = n/2 (2a + (n-1)d

Answer 2

QUESTION:

If the sum of x terms of an A.P.

is 2n² + 5n , then it's nth term

ANSWER:

Given

sum of n terms in an A.P is 2n² + 5n

TO FIND:

nth term of A.P

Sum of n terms of A.P = 2n² + 5n

Take n = 1

We get

2(1)² + 5(1)

S1 = 2 + 5 = 7

If we put n = 1 we get first term because n means no. of terms

first term ( a ) = 7

Take n = 2

S2 = 2(2)² + 5(2)

S2 = 8 + 10 = 18

That means sum of first two terms = 18 = a + t2

Here a = 7

7 + t2 = 18

second term (t2) = 11

Using

Common difference ( d ) = t2 - t1(a)

d = 11 - 7

d = 4

Then,

A.P = 7 , 11 , 15 , .......... n terms

Using

tn = a + (n - 1)d

tn = 7 + (n - 1)4

tn = 7 + 4n - 4

tn = 3 + 4n

So, number of terms ( tn ) = 3 + 4n

More information:

→ Sn = n/2[2a + (n - 1)d]

→ tn = a + (n - 1)d

→ d = t2 - t1

→ a = first term

→ t(n-1) = General term

→ tn = last term

Concept Used :

→ Progressions