Class 10 Maths Extra Questions for Arithmetic Progression

Question 1
The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last terms to the product of the two middle terms is 7 : 15. Find the numbers. Solution

Let the four consecutive numbers in AP be (a - 3d), (a - d), (a + d) and (a + 3d)
According to the question.
$a-3d + a - d + a + d + a + 3d = 32
$
$4a = 32$
$a = 8$ ......(1)
Now,
$\frac {(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac {7}{15}$
$15(a^2 - 9d^2) = 7(a^2 - d^2)
$
$15a^2 - 135d^2 = 7a^2 - 7d^2$
$8a^2 = 128d^2$
Putting the value of a = 8 in above we get.
$
8(8)^2 = 128d^2$
$d^2 = 4$
$d=\pm 2$

Putting a=8, d=+2
the four consecutive numbers are
2, 6, 10 and 14
Putting a=8 ,d=-2 ,we get
the four consecutive numbers are
14, 10, 6 and 2

Question 2
Find the sum of 3 digit numbers which are not divisible by 7? Solution

Sum of all three digit numbers which are not divisible by 7
= Sum of all three digit numbers - Sum of all three digit numbers which are divisible by 7

Now,lets find each of these separately
Sum of all three digit numbers
= 100 + 101 + 102 +...... + 999
Now this series is a AP with first term as 100, common difference =1 and n=900
$S_1 = \frac {n}{2} [a_1 + a_n] = \frac {900}{2} [100 + 999]= 494500$

Sum of all three digit numbers which are divisible by 7
= 105 + 112 + ..... + 994

Now this series is a AP with first term as 105, common difference=7, last term =994, n=?

Now from nth term formula
$a_n= a_1 + (n-1) \times d$
$994 = 105 + (n - 1) \times 7$
$7(n - 1) = 889$
$n = 128$

So, sum of all three digit numbers which are divisible by 7
$S_2 = \frac {n}{2} [a_1 + a_n] = \frac {128}{2} [105 + 994]= 70336$

Therefore,
Sum of all three digit numbers which are not divisible by 7 = $S_1 - S_2 = 494500 - 70336 = 424164$

Question 3
Find the
(i) sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
(ii) sum of those integers from 1 to 500 which are multiples of 2 or 5 Solution

i. LCM of 2 and 5 =10.
Now all those integers which are multiples of 10 are also the multiples of 2 and 5.
So number between 1 and 500 will be given as
10, 20, 30, ...., 490
Sum = 10 + 20 + 30..... + 490

Now this series is a AP with first term as 10, common difference=10, last term =490, n=?

Now from nth term formula
$a_n= a_1 + (n-1) \times d$
$490 = 10 + (n – 1) \times 10$
$n = 49$
So, sum of all three digit numbers which are divisible by 7
$S_{49} = \frac {n}{2} [a_1 + a_n] = \frac {49}{2} [10 + 490]= 12250$

ii.
This question can be divided as
S= Sum of number multiples of 2 + Sum of number multiples of 5 - Sum of number multiples of 2 and 5 both
S=Sum of number multiples of 2 + Sum of number multiples of 5 - Sum of number multiples of 10
$S= S_2 + S_5 - S_{10}$

Now Sum of number multiples of 2
$S_2 =2 + 4 + 6....+ 500$
This is a AP with first term =2 and common difference =2,last term=500
Number of term can found from nth term formula
$a_n= a_1 + (n-1) \times d$
$500 = 2 + (n-1) \times 2$
n=250
Then,
$S_2 =2 + 4 + 6....+ 500 = \frac {n}{2} [a_1 + a_n] = \frac {250}{2} [2 + 500]=62750$

Now Sum of number multiples of 5
$S_5 =5 + 10 + 15....+ 500$
This is a AP with first term =5 and common difference =5,last term=500
Number of term can found from nth term formula
$a_n= a_1 + (n-1) \times d$
$500 = 5 + (n-1) \times 5$
n=100
Then,
$S_5 =5 + 10 + 15....+ 500= \frac {n}{2} [a_1 + a_n] = \frac {100}{2} [5 + 500]=25250$

Now Sum of number multiples of 10
$S_{10} =10 + 20 + 35....+ 500$
This is a AP with first term =10 and common difference =10,last term=500
Number of term can found from nth term formula
$a_n= a_1 + (n-1) \times d$
$500 = 10 + (n-1) \times 10$
n=50
Then,
$S_5 =10 + 20 + 30....+ 500= \frac {n}{2} [a_1 + a_n] = \frac {50}{2} [10 + 500]=12750$
Therefore,
$S= S_2 + S_5 - S_{10}$
$S= 62750+25250-12750$
$S= 75750$

Question 4
Find the sum of all three digit numbers which leave the remainder 3 when divided by 5 Solution

The smallest 3 digit no. = 100 and greatest 3 digit number is 999

Since 100 is divisible by 5,adding 3 on 100 will provide the number which leave the remainder 3 when divided by 5

So the smallest three digit number which is divisible by 5 and gives reminder 3= 103

The largest 3 digit no ,
999/5 gives reminder 4 ,So subtracting 1 will the number which leave the remainder 3 when divided by 5

So the Largest three digit number which is divisible by 5 and gives reminder 3= 998

Similarly we can find other numbers, the number will be given as
103,108,111,...998

Now $S= 103+ 108 + 111... + 998$
This is a AP with first term =103 and common difference =5,last term=998
Number of term can found from nth term formula
$a_n= a_1 + (n-1) \times d$
$998 = 103 + (n-1) \times 5$
n=180

Question 5
In an AP, if $S_n = 3n^2+ 5n$ and a_{k} = 164, find the value of k. Solution

Given: $S_n = 3n^2+ 5n$
$S_1 = 3(1)^2 + 5(1)
= 3 + 5=8$
Therefore
$a_1 = 8$
$S_2 = 3(2)^2 + 5(2)
= 12 + 10
= 22
$
Therefore
$a_1+ a_2 =22$
$
8 + a_2 = 22$
$a_2 = 14$
So, the AP is
8 ,22......
So, common difference=14
Now from nth term formula
$a_n= a_1 + (n-1) \times d$
$164 = 8 + (k-1)14$
$k = 27
$

Question 6
If $S_n$ denotes the sum of first n terms of an AP, prove that
$S_{12}= 3(S_8- S_4)$ Solution

Let a is the first term of A.P and d is the common difference
$
S_n=\frac {n}{2} {2a+(n-1)d}$
$S_{12}=\frac {12}{2} {2a+(12-1) d}=12a+66d
$
$_S8=\frac {8}{2} {2a+7d}=8a+28d$
$
S_4=\frac {4}{2} {2a+3d}=4a+6d$

Question 8
Find the 20th term from the end of the AP 3, 8, 13......253 Question 9
The sum of the first n terms of an A.P whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is - 30 and the common difference is 8. Find n. Solution

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