Question

find the sum of the first 14 terms for a sequence starting with 2,ending with 120,common difference is 2

Answer 1

sequence is in AP , because here mentioned about common difference .

first term, a = 2
last term , Tn = 120
common difference , d = 2
use formula , Tn = a + (n - 1)d
120 = 2 + (n -1)×2
118/2 = n - 1
59+1 = n
n = 60

but we have to find sum of first 14 terms
so, $T_{14}=2+(14-1)\times2=28$

now, sum of first 14 terms = $\frac{14}{2}[a+T_{14}]$
= 7(2 + 28)
= 7 × 30
= 210

[ note :- sum of n terms = n/2[first term + nth term]]

Answer 2

Find the AP:

an = a1 + (n - 1)d

Given that the first term is 2 and the common difference is 2.

an = 2 + (n - 1)2

an = 2 + 2n - 2

an = 2n

Find the 14th term:

an = 2n

when n = 14

a14 = 2(14) = 28

Find the sum of the first 14 terms:

sn = n/2 (a1 +an)

s14 = 14/2 (2 + 28) = 210

Answer: The sum of the first 14th term is 210.