Question

Q4. How many terms of the A.P. -6, -11/2, -5, ... are needed to give the sum -25?​

Answer 1

Step-by-step explanation:

Given :-

-6, -11/2, -5, ...

To find :-

How many terms of the A.P. -6, -11/2, -5, ... are needed to give the sum -25?

Solution:-

Given that :

-6 , -11/2 , -5 , ... are in the AP

First term = (a) = -6

Common difference = (d)= an - a(n-1)

= (-11/2)-(-6)

=(-11/2)+6

=(-11+12)/2

d = 1/2

Let the number of terms are needed to give the sum -25 be "n"

We know that

Sum of first n terms in an AP = Sn

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

= Sn = -25

=> (n/2)[2(-6)+(n-1)(1/2)] = -25

=> (n/2)[-12+(n-1)(1/2)] = -25

=> (n/2)[{-24+n-1}/2] = -25

=> (n/2)[-25+n]/2 = -25

=> (n)(n-25)/4 = -25

=> n(n-25) = -25×4

=> n(n-25) = -100

=> n^2-25n = -100

=> n^2-25n+100 = 0

=> n^2-5n-20n+100 = 0

=> n(n-5)-20(n-5) = 0

=> (n-5)(n-20) = 0

=> n-5 = 0 or n-20 = 0

= > n = 5 or n=20

Therefore,n= 5 or 20

Answer:-

Required number of terms in the AP is 5 and 20

Check:-

If n = 5 then

S5 = (5/2)[2(-6)+(5-1)](1/2)

=> (5/2)[-12+(4/2)]

=> (5/2)[-12+2]

=> (5/2)(-10)

=> -50/2

=> -25

S5 = -25

If n = 20 then

S20 = (20/2)[2(-6)+(20-1)(1/2)]

=> (10)[-12+19(1/2)]

=> (10)(-24+19)/2

=> (10)(-5/2)

=> -50/2

=> -25

Verified the given relations in the problem

They are true for n = 5 and 20

Used formulae:-

• Sum of first n terms in an AP =
• Sn= (n/2)[2a+(n-1)d]
• a = First term
• d = Common difference
• d = an-a(n-1)
• an = nth term
• a(n-1)= (n-1)th term

Answer 2

Answer :.

it is the answer

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