Q4. How many terms of the A.P. -6, -11/2, -5, … are needed to give the sum -25? About the author Caroline
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 Reply
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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:–