Question

find the sum of the first 51 terms of an ap .whose 2 and 3rd terms are 14 and 18​

Answer 1

Answer:

Sum of 51 terms is 5610

Step-by-step explanation:

Let first term be 'a' and common difference be 'd'

Then

2nd term : 14=a+(2-1)d 14=a+d

3rd term : 18=a+(3-1)d 18=a+2d=a+d+d

On substituting value of (a+d) from 1st equation into 2nd equation

18=14+d d=4

From 1st equation

14=a+4 a=10

Now, sum if n terms:

Sₙ = (n/2)×[2a+(n-1)d]

Here n=51 a=10 and d=4

So,

S₅₁ = (51/2)×[(2×10) + (51-1)×4]

= (51/2)×[20+ 50×4]

= (51/2)×(220)

= (51×110)

S₅₁ = 5610

Answer 2

ANSWER :

Let the first term be 'a' and the common difference be 'd'.

2nd term:-

T2 = a+d

14 = a+d_____eq1

3rd term:-

T3= a+2d

18 = a+2d_____eq2

Subtracting eq1 from eq2, we get

d = (18-14)=4

Now, putting the value of d in eq 1, we have (a+4)=14, so a = 10.

To Find :

Sum of 51 terms of the Arithmetic Progression.

Calculation :

$Sn = \frac{n}{2} (2a + (n - 1)d)$

$S51 = \frac{51}{2} (2 \times 10 + (51 - 1)4)$

$S51 = \frac{51}{2} (20 + 50 \times 4)$

$S51 = \frac{51}{2} (220) \\ S51 = 5610$