Question

If the first term of an ap is 2 and the sum of the first five terms is equal to one fourth of the sum of the next five terms then the sum of the first 30 terms

Answer 1

Answer:

-255

Step-by-step explanation:

Nice one. First of all let's write all given situations. For this A.P. :-

a = 2 and S5 = (S10-S5)/4

Lete explain the second one. The sum of first 5 terms is S5 and sum of first 10 terms is S10. So, sum of next 5 terms ( i.e sum of 6th to 10th term ) is S10 - S5.

So, from this equation we get 5S5 = S10

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

So, S5 = 5/2( 4 + 4d )

or S5 = 2.5×4(1+d) = 10(1+d)

And S10 = 10/2( 4+ 9d ) = 5(4+9d )

So, 50(1+d) = 5(4+9d)

10 + 10d = 4 + 9d

d = -6

So, Sum of first 30 terms = S30

= 30/2(4+29d)

=15(4–174)

= 15(-170)

= -255

So, required answer is -255

Answer 2

Answer:

-2550

Step-by-step explanation:

Given, a = 2, let common difference be d.

So, Sum of first 5 terms, $\sf S_5$ = n/2{2a + (n - 1)d}

= 5/2{2(2) + (5 - 1)d}

= 5/2{4 + 4d}

= 10 + 10d

Now, the next five terms would be a + 5d (the 6th term), a + 6d.... a + 9d (the tenth term)

Now, sum = n/2(a + l),

where, l is the last term

So here, sum = 5/2(a + 5d + a + 9d)

→ sum = 5/2(2a + 14d)

→ sum = 5a + 35d

Now, one fourth of this is equal to the sum of first five terms

→ 10 + 10d = (5a + 35d)/4

→ 4(10 + 10d) = 5a + 35d

→ 40 + 40d = 5a + 35d

→ 40 + 40d - 35d = 5a

→ 5a = 40 + 5d

Now, a is 2

→ 10 = 40 + 5d

→ 5d = 10 - 40

→ d = -30/5

→ d = -6

So, sum of first 30 terms would be

n/2{2a + (n - 1)d}

$\sf\frac{30}{2}\{2(2) + (30 - 1)(-6)\} \\ \\ \implies 15(4 -174)\\ \\ \implies 15(-170) \\ \\ \implies -2550$

Answer :- (- 2250 )