Question

If nth term of an AP is 4 + 2n, then find the sum of first 5 terms

Answer 1

Answer:

nth term is 4

a+(n-1)d=4

a+(n-1)2=4 ,as d=4 given

a+2n-2=4

a+2n=6

a=6-2n

the sum of n terns is

n/2[2a+(n-1)2]= -14

n(a+n-1)= -14

as we know a value substitute it in this equation

n(6-2n+n-1)= -14

n(5-n)= -14

n^2-5n-14=0

factorise it

n^2-7n-5n-14=0

n(n-7)+2(n-7)=0

(n-7)(n-2)=0

n=7 or n=-2

then value of a is

10 or-8

hope it helps u

Step-by-step explanation:

Answer 2

Answer:

$\red\bigstar$ The sum of first 5 terms is 50 .

   SOLUTION    

Given:- The nth Term of the AP is 4 + 2n .

Let us put , n= 1

a₁ = 4 + 2×1 = 4 + 2 = 6

Let us put, n = 2

a₂ = 4 + 2 × 2 = 4 +4 = 8

Let us put, n = 3

a₃ = 4 + 2 × 3 = 4 + 6 = 10

The AP is 6,8,10,...

Common Difference (d) = a₂ - a₁ = 8 - 6 = 2

SUM OF FIRST FIVE TERMS OF THE AP

$\bigstar$ Sn = n/2[2a + (n − 1) × d]

S₅ = $\dfrac{5}{2} (2*6+(5-1)*2)$

     =$\dfrac{5}{2} (12 + 4*2)$

     = $\dfrac{5}{2} * 20$

S₅  = 5×10

$\red\bigstar$ Sum of First five terms of the AP = 50.