$a_n = -4n + 15$
$a_k = -4k + 15$
$a_{k+1} = -4(k+1) + 15$
Now
$a_{k+1} - a_k=-4(k+1) + 15 -[-4k + 15]=-4$
Since difference between two terms constant.It is a AP
$a_{15} = -4(15) + 15=-45$
a_{n}=6n + 11
a_{k} = 6k + 11
a_{k+1} = 6(k+1) + 11
Now
a_{k+1} - a_{k}=6(k+1) + 11 -[6k + 11]=6
a_{8} = 31
a +(8-1)d = 31
a + 7d = 31 -- (i)
a_{15} =16 + a_{11}
a + 14d = 16 + a +10d
14d = 16 + 10d
4d = 16
d = 4
Now putting d = 4 in eq. (i) we get
a + 7(4) = 31
a + 28 = 31
a = 3
So AP
3,7,11,15...
Let n th term be 130 more than the 31st term of the A.P.
First term of A.P. = 5
Common difference = 15 - 5 = 10
a_{n} = 130 + a_{31}
5 + (n - 1) X 10 = 130 + 5 + (31 - 1) X 10
10 (n - 1) = 430
n = 44
Thus, 44th term of the A.P is 130 more than the 31st term.
Similar question as above.
Answer is 65th term is 132 more than its 54th term
$a_n=a + (n-1)d$
For Ist AP
$a_{100x}=a_x + (100-1)d$
For 2nd AP
$a_{100y}=a_y + (100-1)d$
So difference
$a_x-a_y=a_{100x}- a_{100y}$
$a_x-a_y=111222333$
This difference will remain in all the terms
So ,answer is 111222333 only
$41=a + 9d$
$73=a+ 17d$
Solving these
a=5,d=4
$a_{26} =a + 25d= 105$
(m+1)th term= a + (m+1-1) d = a + m d
(n +1) th term = a+ (n+1-1) d = a + n d
now given condition is
a+ md = 2 ( a + n d ) ---- (1)
Now
(3m + 1) th term = a + (3m+1-1) d= a + 3m d -- (2)
(m+n+1) th term = a + (m+n +1 -1) d = a + (m+n) d ---(3)
Now
(3m+1)th term= a +3md
=a +md + 2md
Now we have a+md = 2 (a + nd ) from equation 1
=2(a+nd) + 2md
= 2(a + nd + md)
= 2(a +(m+n)d )
=2 (m+n+1) th ( from (3) )
Hence proved
For 1st AP
a=9,d=-2
$a_{nx}= 9 - (n-1)2$
For 2nd AP
a=15,d=-3
$a_{ny}= 15 - (n-1)3$
Now
$9 - 2(n-1)=15 - 3(n-1)$
n=7
$a+ 5d=12$
$a+ 7d=22$
Solving these
d=5,a=-13
$a_2= -13 + 5= -8$
$a_n= -13 + (n-1)5= 5n-18$
$a + 3d + a + 7d=24$ or $2a + 10d=24$
$a+ 5d+ a + 9d=34$ or $2a + 14d =34$
Solving these
a=-1/2 ,d=5/2
According to the question
$m(a_m) = n(a_n)$
Now for a,d as first term and common difference of the AP,nth term is defined as ,
$a_n= a + (n-1)d$
So
$m[a + (m-1)d] = n[a + (n-1)d]$
$[ma + (m^2 - m)d]= [na + (n^2 - n)d]$
$[ma + (m^2d- md)]= [na + (n^2d- nd)]$
$[ma-na] + [m^2d- n^2d] + [nd-md] = 0$
$a[m-n] + d[m^2 - n^2] + d[n-m] = 0$
Now we know that $x^2 - y^2 = (x+y) (x-y)$
$a[m-n] + d[(m+n) (m-n)] - d[m - n] = 0$
Divide the above equation with (m-n) ,We get :
$a + d(m+n) - d = 0$
$a + [ (m+n) - 1 ] d = 0$
So $a_{m+n} = 0$
We can solve these question in three ways
a. Find the Ist,2nd ,3rd and 4th term and check for the common difference. If same then in AP
b. Find d as $d=a_n - a_{n-1}$. Now if d does not depend on n, then it is in AP
c. We know that in an AP $a_n=a + (n-1)d$. We observe that $a_n$ is a linear polynomial in n. So if the expression is not linear,it is not an AP
i. $a_n=2n -3$
$a_{n-1} =2(n-1) -3=2n-5$
$d=a_n - a_{n-1}= 2n -3 - 2n + 5= 2$
So this is an nth term of A.P
ii. $a_n =3n^2+5$
This is not an linear polynomial,so it is not an nth term of A.P
iii. $a_n=1+n+n^2$
This is not an linear polynomial,so it is not an nth term of A.P
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 = $S_1 - S_2 = 494500 - 70336 = 424164$
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$
This Class 10 Maths Important Questions for Arithmetic Progression with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.