Question

Let nC2, (n^2-3n+36)/2 and n+1C2 be the first three terms of arithmetic sequence. Find the sum of first 25 terms of that sequence.​

Answer 1

Step-by-step explanation:

Given :-

nC2, (n²-3n+36)/2 and (n+1)C2 be the first three terms of arithmetic sequence.

To find :-

Find the sum of first 25 terms of that sequence?

Solution :-

Given that :

nC2, (n²-3n+36)/2 and (n+1)C2 be the first three terms of AP

we know that

nCr = n!/[(r)!(n-r!)]

n ! = (n)(n-1)(n-2)..(1)

=>nC2 = [n!/(2!×(n-2)!]

=> nC2 = (n)(n-1)(n-2)!/2(n-2)!

=> nC2= n(n-1)/2

nC2 = (n²-n)/2 ------------(1)

and

(n+1)C2 = [(n+1)!/(2!×(n+1-2)!]

=> (n+1)C2 =[(n+1)!/(2!×(n-1)!]

=> (n+1)C2 = [(n+1)!/(2!×(n-1)(n-2)!]

=>(n+1)C2=[(n+1)(n)(n-1)(n-2)!/(2!×(n-1)(n-2)!]

=> (n+1)C2 = (n+1)(n)/2

=> (n+1)C2 = (n²+n)/2----------(2)

We know that

If a , b and c are the first three terms in an APthen 2b = a+c

We have

a = nC2 = (n²-n)/2

b= (n²-3n+36)/2

c = (n+1)C2 = (n²+n)/2

=> 2[(n^2-3n+36)/2] = nC2 + (n+1)C2

=> n²-3n+36 = nC2 + (n+1)C2

From (1)&(2)

=> n²-3n+36 = [(n²-n)/2]+[(n²+n)/2]

=> n²-3n+36 = (n²-n+n²+n)/2

=> n²-3n+36 = (n²+n²)/2

=> n²-3n+36 = 2n²/2

=> n²-3n+36 = n²

=> n²-3n+36-n² = 0

=> -3n+36 = 0

=> -3n = -36

=> 3n = 36

=> n = 36/3

=> n = 12

Therefore, n = 12

On Substituting the value of n in the given three terms then they will be

First term = (n²-n)/2

=>a1= (12²-12)/2

=>a1 = (144-12)/2

=>a1 = 132/2

=>a1= 66

Second term= (n²-3n+36)/2

=>a2 = (12²-3(12)+36))/2

=>a2 = (144-36+36)/2

=> a2 = 144/2

=> a2 = 72

Third term = (n²+n)/2

=> a3 = (12²+12)/2

=>a3 = (144+12)/2

=> a3 = 156/2

=>a3= 78

The first three terms of the AP = 66,72,78

Common difference = 72-66 = 6

Finding Sum of 25 terms :-

We know that

Sum of the first n terms in an AP = Sn

=(n/2)[2a+(n-1)d]

We have a = 66 and d = 6 and n = 25 then

The sum of first 25th terms in the AP

=> S25

=> (25/2)[2(66)+(25-1)(6)]

=> S 25 = (25/2)[132+24(6)]

=> S 25 = (25/2)[132+144]

=> S 25 = (25/2)(276)

=> S 25 = (25×276)/2

=> S 25 = 25×138

=> S 25 = 3450

Therefore, S 25 = 3450

Answer :-

The sum of first 25 terms in the given AP is 3450

Used formulae:-

• nCr = n!/[(r)!(n-r!)]

• n! = (n)(n-1)(n-2)...1

• Sum of the first n terms in an AP = Sn =(n/2)[2a+(n-1)d]

• a = First term

• d = Common difference

• n = number of terms

• If a , b and c are the first three terms in an APthen 2b = a+c or b= (a+c)/2