Question

heyy...guys ......can u plzz solve this sum?

Answer 1

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

Given that the first term is 100.

= > a = 100.

We know that sum of n terms of an AP sn = (n/2)(2a + (n - 1) * d)

Sum of 1st 6 terms:

s6 = (6/2)(2(100) + (6 - 1) * d

     = 3(200 + 5d)
 
     = 600 + 15d.



Sum of next 6 terms:

s12 = (12/2)(2(100) + (12 - 1) * d) - 600 + 15d

      = (6)(200 + 11d) - 600 + 15d

      = 1200 + 66d - 600 + 15d
 
      = 600 + 51d.


Given that sum of 1st 6 terms is 5 times the sum of the next 6 terms.

= > 600 + 15d = 5(600 + 51d)

= > 600 + 15d = 3000 + 255d

= > 600 + 15d - 3000 = 255d 

= > -2400 = 240d

= > d = -10.



Therefore the common difference is d = -10.


Hope this helps!

Answer 2

Let the first term AP be a and common difference be d ,
Given ,
a= 100

Sum of first 6 terms = 6/2{2a +(6-1)d}
S6= 3{2a 5d}
S6 = 6a +15d

Now,
Sum of next 6 terms = sum of 12 terms - sum of First 6 terms
→S12 = 12/2{2a + (12-1)d} - { 6a +15d}

→ S12 = 6{ 2a + 11d} - {6a +15d}

→S12 = 12a +66d -6a -15d

S12 = 6a + 51d

According to question,
S6 = 5*S12

6a +15d = 5×{ 6a + 51d}

→3{2a +5d} = 30a + 255d

→6a +15d = 30a + 255d
→24a = - 240d
Now putting a= 100
→24*100 = -240d
d= -10