Class 10 Maths Extra Questions for Arithmetic Progression

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

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 = $S_1 - S_2 = 494500 - 70336 = 424164$

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,
$S= S_2 + S_5 - S_{10}$
$S= 62750+25250-12750$
$S= 75750$

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

Therefore
$S= 103+ 108 + 111... + 998 = \frac {n}{2} [a_1 + a_n] = \frac {180}{2} [103 + 998]=99090$

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 $

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$

Now we have to prove
$S_{12}= 3(S_8- S_4)$

Taking LHS

$LHS=S_{12}=12a+66d$

Taking RHS
$RHS=3 (S_8-S_4)=3 (8a+28d-4a-6d)=12a+66d= LHS$

Hence proved

Let a is the first term of A.P and d is the common difference
$a_p= a+ (p-1) d=x$
$a_q = a + (q-1)d=y$
$a_r = a + (r-1)d=z$

Substituting these values of x,y and z in below equation

$x(q-r) + y(r-p) + z(p-q) $
$={a+ (p-1) d}(q-r) + {a+ (q-1) d} (r-p) + {a+ (r-1) d} (p-q)
$=a {(q-r) + (r-p) + (p-q)} + d {(p-1)(q-r)+ (r-1) (r-p) + (r-1) (p-q)}
$= a.0 + d{p(q-r) + q(r-p) + r (p-q)- (q-r) - (r-p)-(p-q)}
$= a.0 + d.0 = 0$
Hence proved

$S_n=\frac {n}{2}(2a + (n-1)d)$
Given, a=8 and d=20
$S_n=\frac {n}{2}(2(8) + (n-1)20)=n(10n-2)$

When a=-30 and d=8
$S_{2n}=\frac {2n}{2}(2(-30) + (2n-1)8)=n(-68 + 16n)$
According to question
$S_n= S_{2n}$
$n(10n-2)=n(-68 + 16n)$
n=11

Given
$t_1 = k^2 + 4k + 8$
$t_2 = 2k^2 + 3k + 6$  
$t_3 = 3k^2 + 4k + 4$
Now since they are in A.P
$ t_2 - t_1 = t_3 - t_2$
$ 2k^2 + 3k + 6 - (k^2 + 4k + 8) = (3k^2 + 4k + 4) - (2k^2 + 3k + 6)$
$ 2k^2 + 3k + 6 - k^2 - 4k - 8 = 3k^2 + 4k + 4 - 2k^2 - 3k - 6$
$ k = 0$