Question

Q.In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.
Q.2)How many terms of the A.P. -6, -11/2, -5, …. are needed to give the sum –25?

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Answer 1

Question 01

$\sf \: Sum \: of \: first \: n \: terms = \\ \tt \red{S_n=\frac{n}{2} (2a+(n-1)d)}$

Where

• n is the number of terms
• a is the first term
• d is the common difference

We are given that the sum of the first five terms is one- fourth of the sum of the next- five terms.

$\tt \: S_5=\frac{1}{4}(S_{10}-S_5) \\ \\ \tt \: \frac{5}{2} (2(2)+(5-1)d)=\frac{1}{4}(\frac{10}{2})\\ \\ \tt \: (2(2)+(10-1)d)-\frac{5}{2}\\ \\ \tt \: (2(2)+(5-1)d))\\ \\ \tt \: 20+20d=\frac{1}{4}(40+90d-(20+20d))\\ \\ \tt \: 80+80d=40+90d-20-20d\\ \\ \tt \: 80+80d=20+70d\\ \\ \tt \: 10d=-60\\ \\ \tt \: d=-6$

Formula of nth term in A.P. =

$\tt \: a_n=a+(n-1)d$

Substitute n 20

$\tt \: a_{20}=2+(20-1)(-6)$

$\tt \: a_{20} = - 120$

___________________________

Question 02

• 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

Answer 2

Answer 01

$\begin{gathered} \sf \: Sum \: of \: first \: n \: terms = \\ \tt \red{S_n=\frac{n}{2} (2a+(n-1)d)}\end{gathered}$

Where

• n is the number of terms
• a is the first term
• d is the common difference

We are given that the sum of the first five terms is one- fourth of the sum of the next- five terms.

$\begin{gathered} \tt \: S_5=\frac{1}{4}(S_{10}-S_5) \\ \\ \tt \: \frac{5}{2} (2(2)+(5-1)d)=\frac{1}{4}(\frac{10}{2})\\ \\ \tt \: (2(2)+(10-1)d)-\frac{5}{2}\\ \\ \tt \: (2(2)+(5-1)d))\\ \\ \tt \: 20+20d=\frac{1}{4}(40+90d-(20+20d))\\ \\ \tt \: 80+80d=40+90d-20-20d\\ \\ \tt \: 80+80d=20+70d\\ \\ \tt \: 10d=-60\\ \\ \tt \: d=-6\end{gathered}$

Formula of nth term in A.P. =

$\tt \: a_n=a+(n-1)d$

Substitute n 20

$\tt \: a_{20}=2+(20-1)(-6) \\ \tt \: a_{20} = - 120$

_________________________

Solution 02:

First term = (a) = -6

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

$\sf \: = (-11/2)-(-6) \\ \sf \:=(-11/2)+6\\ \sf \: =(-11+12)/2\\ \sf \: d = 1/2$

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

Sum of first n terms in an AP = Sn

$\sf \: = (n/2)[2a+(n-1)d] \\ \sf \: = Sn = -25\\ \sf \: => (n/2)[2(-6)+(n-1)(1/2)] = -25\\ \sf \: => (n/2)[-12+(n-1)(1/2)] = -25\\ \sf \: => (n/2)[{-24+n-1}/2] = -25\\ \sf \: => (n/2)[-25+n]/2 = -25\\ \sf \: => (n)(n-25)/4 = -25\\ \sf \: => n(n-25) = -25×4\\ \sf \: => n(n-25) = -100\\ \sf \: => n^2-25n = -100\\ \sf \: => n^2-25n+100 = 0\\ \sf \: => n^2-5n-20n+100 = 0\\ \sf \: => n(n-5)-20(n-5) = 0\\ \sf \: => (n-5)(n-20) = 0\\ \sf \: => n-5 = 0 or n-20 = 0$

Therefore,n= 5 or 20

Answer:-

Required number of terms in the AP is 5 and 20