Question

If 3rd term of an A.P. is 20 and 10th term is 55, then find the sum of first fifty terms.

Answer 1

Answer:

6625

Step-by-step explanation:

In APs, nth term = a + (n - 1)d, where a is first term and d is common difference.

Let first term of this AP be a and common difference be d.

= > 3rd term = 20

= > a + 2d = 20

= > a = 20 - 2d ... (1)

Whereas,

= > 10th term = 55

= > a + 9d = 55

= > 20 - 2d + 9d = 55 { a = 20 - 2d }

= > 7d = 55 - 20

= > 7d = 35

= > d = 5

Thus, a = 20 - 2d = 20 - 2(5) = 20 - 10

a = 10

Sum of n terms is given be (n/2) { 2a + (n - 1)d }

So,

Sum of first fifty terms:

= > (50/2) { 2(10) + (50 - 1)5 }

= > 25{ 20 + 245 }

= > 6625

Answer 2

Answer:

Step-by-step explanation:

T3=20

So, a+2d=20 ————-i

T10=55

So, a+9d=55———ii

Now by subtracting equation i from ii we get,

a+9d-(a+2d)=55-20

9d-2d=35

7d=35

d=5

So, a=20-10 ( substituting in equation i )

a=10

Now using Sn=n/2(2a+(n-1)d)

n=50

Sn=50/2(2*10+(50-1)*5)

Sn=6625

Therefore the sum of the first 50 terms is 6625