Question

If in an A.P. a1+a2+a6+a7+a11+a12=300; then sum of first twelve terms of that A.P. is

Answer 1

The sum of first twelve terms of this A.P is 600.

S12 = 600

a1 + a2 + a6 + a7 + a11 + a12 = 300

Since an = a + (n - 1)d

Therefore, applying the formula for above values we will get

a1 = a

a2 = a + (2 - 1) d = a + d

a6 = a + (6 - 1) d = a + 5d

a7 = a + (7 - 1) d = a + 6d

a11 = a + (11 - 1) d = a + 10d

a12 = a + (12 - 1) d = a + 11d

Now putting all the values in the given equation we will get

a + a + d + a + 5d + a + 6d + a + 10d + a + 11d = 300

6a + 33d = 300

3 (2a + 11d) = 300

2a + 11d = 100 ............ eq. 1

We know that,

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

S12 = 12/2 [2a + (12 - 1)d]

S12 = 6 [2a + 11d] ........... eq. 2

Putting the value of eq. 1 in eq. 2 we get

S12 = 6 (100)

S12 = 600