Arithmetic progression
Leave a Comment	/ Class 11, RD sharma / By Tanishk
EXERCISE 19.2
Find :
(i) 10 th term of the A.P.
Sol :
d = 4 – 1 = 3
(ii) 18 th term of the A.P.
We have :
(iii) n th term of the A.P.
a = 13
d = 8 – 13 = -5
If the sequence is an A.P. show that
Let the sequence be an A.P. with the first term being A and the common difference being D .
To prove:
From (i) and (ii), we get:
LHS = RHS
Hence proved
(i) Which term of the A.P. is 248 ?
d = (8-3) = 5
Hence, 248 is the 50 th term of the given A.P.
(ii) Which term of the A.P. is 0 ?
d = (80-84) = -4
Hence , 0 is the 22nd term of given A.P
(iii) Which term of the A.P. is 254 ?
Here , we have
d = (9 – 4) = 5
Hence, 254 is the 51 st term of the given A.P.
(i) Is 68 a term of the A.P. ?
d= (10 – 7) = 3
Since n is not a natural number.So, 68 is not a term of the given A.P.
(ii) Is 302 a term of the A.P. ?
a= 3
Since n is not a natural number. So ,302 is not a term of the given A.P.
(i) Which term of the sequence is the first negative term?
This is an A.P
Let the first negative term be
Then , we have :
Thus, the 34 th term is the first negative term of the given A.P.
(ii) Which term of the sequence is (a) purely real (b) purely imaginary?
Let the real term be
Let nth term be purely real
So, 5 th term is purely real.
Let n th term be purely imaginary. Then,
So, 13 th term is purely imaginary.
REASON : A complex number is a number which can be represented in the form of 'z=a+ib' , where a & b are real number and i is an imaginary number, (which is a square root of -1). If a=0 then z is purely imaginary number but if b=0 then z is purely real number.
(i) How many terms are there in the A.P.
d = (10 – 7) =3
Let there be n terms in the given A.P.
Thus, there are 13 terms in the given A.P.
(ii) How many terms are there in the A.P. ?
a = -1
The first term of an A.P. is 5 , the common difference is 3 and the last term is 80 ; find the number of
d=3
a_{n}=80 a_{n}=a+(n-1) d \Rightarrow 80=5+(n-1) 3 \Rightarrow 75=(n-1) 3 \Rightarrow 25=(n-1) \Rightarrow 26=n
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Thus , there are 26 terms in the given A.P
<strong><span style="background-color: #ffff00;">QUESTION 8</span></strong>
The 6 th and 17 th terms of an A.P. are 19 and 41 respectively, find the 40 th term.
Given :
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Missing $ inserted.
a_{6}=19 \Rightarrow a+(6-1) d=19\left[a_{n}=a+(n-1) d\right] \Rightarrow a+5 d=19 \quad \ldots(1) a_{17}=41 \Rightarrow a+(17-1) d=41 \quad\left[a_{n}=a+(n-1) d\right] \Rightarrow a+16 d=41 \quad \ldots(2) 16 d-5 d=41-19 \Rightarrow 11 d=22 \Rightarrow d=2 d=2 a+5 \times 2=19 \Rightarrow a=19-10 \Rightarrow a=9
We know:
a_{40}=a+(40-1) d \quad\left[a_{n}=a+(n-1) d\right] =a+39 d =9+39 \times 2
=9+78 =87
If 9 th term of an A.P. is zero, prove that its 29 th term is double the 19 th term.
a_{29}=2 a_{19}
a_{9}=0 \Rightarrow a+(9-1) d=0 \quad\left[a_{n}=a+(n-1) d\right] \Rightarrow a+8 d=0 \Rightarrow a=-8 d \ldots(\mathrm{i})
LHS:
a_{19}=a+(19-1) d =a+18 d = -8d+18d = 10d \dots (ii)
RHS :
a_{29}=a+(29-1) d =a+28 d =-8 d+28 d = 20 d \dots (iii)
<strong><span style="background-color: #ffff00;">QUESTION 10</span></strong>
If 10 times the 10 th term of an A.P. is equal to 15 times the 15 th term, show that 25 th term of the A.P. is zero.
10 a_{10}=15 a_{15} \Rightarrow 10[a+(10-1) d]=15[a+(15-1) d] \Rightarrow 10(a+9 d)=15(a+14 d) \Rightarrow 10 a+90 d=15 a+210 d \Rightarrow 0=5 a+120 d \Rightarrow 0=a+24 d \Rightarrow a=-24 d \ldots(\mathrm{i})
To show:
a_{25}=0
Proof :
\Rightarrow \mathrm{LHS} : a_{25}=a+(25-1) d =a+24 d =-24 d+24 d =0=\mathrm{RHS}
Hence , proved
The 10 th and 18 th terms of an A.P. are 41 and 73 respectively. Find 26 th term.
a_{10}=41 \Rightarrow a+(10-1) d=41 \left[a_{n}=a+(n-1) d\right] \Rightarrow a+9 d=41 a_{18}=73 \Rightarrow a+(18-1) d=73\left[a_{n}=a+(n-1) d\right] \Rightarrow a+17 d=73 \Rightarrow 17 d-9 d=73-41 \Rightarrow 8 d=32 \Rightarrow d=4 \quad \ldots(i) a+9 \times 4=41 \Rightarrow a+36=41 \Rightarrow a=5 \quad \ldots(ii)
a_{26}=a+(26-1) d \quad\left[a_{n}=a+(n-1) d\right] \Rightarrow a_{26}=a+25 d \Rightarrow a_{26}=5+25 \times 4from~(i)~and~(ii) \Rightarrow a_{26}=5+100=105
In a certain A.P. the 24 th term is twice the 10 th term. Prove that the 72 nd term is twice the 34 th
a_{24}=2 a_{10} \Rightarrow a+(24-1) d=2[a+(10-1) d] \Rightarrow a+23 d=2(a+9 d) \Rightarrow a+23 d=2 a+18 d \Rightarrow 5 d=a \quad \ldots(\mathrm{i})
: a_{72}=a+(72-1) d \Rightarrow a_{72}=a+71 d \Rightarrow a_{72}=5 d+71 d \Rightarrow a_{72}=76 d
\mathrm{RHS} : 2 a_{34}=2[a+(34-1) d] \Rightarrow 2 a_{34}=2(a+33 d) \Rightarrow 2 a_{34}=2(5 d+33 d) \Rightarrow 2 a_{34}=2(38 d) \Rightarrow 2 a_{34}=76 d \therefore \mathrm{RHS}=\mathrm{LHS}
Hence, proved.
(m+1) (n+1) (3 m+1) (m+n+1)$ th term.