Question

The first term of an AP is 2 and the c.d is 4.is it possible that the sum of any consecutive ten terms of the sequence is 500?​

Answer 1

Answer:

No.

Step-by-step explanation:

In APs, if first term is a and last term is l, sum of n terms is given by (n/2) [ a + l ]

Also, nth term = a + ( n - 1 )d, where d is common difference.

Here,

a = first term = 2

d = common difference = 4

Let any term nth, after 10 terms, here comes ( n + 9 )th term.

So,

nth term = a + ( n - 1 )d

(n+9)th term = a + ( n + 9 - 1 )d = a + ( n + 8 )d

Hence, let 500 be their sum,

= > (10/2) [ a + ( n - 1 )d + a + ( n + 8 )d ] = 500

= > 5[ a + nd - d + a + nd + 8d ] = 500

= > 2a + 2nd + 7d = 100

= > 2(2) + 2n(4) + 7(4) = 100 {a=2;d=4}

= > 4 + 8n + 28 = 100

= > 8n = 68

= > n = 68/8 = 34/4

As n is in fraction, sum of any consecutive 10 terms can't be 500, for this AP.