Question

28. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18
respectively.
S
obot​

Answer 1

Answer:

5610

Step-by-step explanation:

Let first term be a and common difference be d

We have :

a + ( 2 - 1 ) d = 14

a + d = 14

a = 14 - d .... ( i )

a + ( 3 - 1 ) d = 18

a + 2 d = 18

a = 18 - 2 d ... ( ii )

From ( i ) and ( ii )

18 - 2 d = 14 - d

d = 4

a = 14 - d

a = 10

Now :

S₅₁ = 51 / 2 ( 20 + 50 × 4 )

S₅₁ = 51 ( 10 + 100 )

S₅₁ = 5610

Answer 2

Answer:

t2=14 and t3=18

d=4

a=t1=10

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

S51=51/2[2×10+(51-1)4]

=51/2[20+50×4]

=51/2(220)

=51×110

=5610