10th Quadratic Equations:Solving by Completing the Square

CBSE I NCERT 10th Arithmetic Progression Problems with solution

Solved Arithmetic Progression Problems By Jsunil

(1) Determine k so that k+2 , 4k-6 and 3k-2 are the three consecutive terms of an AP.

a₁= k+2

a₂= 4k-6

a₃= 3k-2

a₂-a₁= a₃- a₂

4k-6-(k+2) = 3k-2-(4k-6)

4k -6-k-2 = 3k-2-4k+6

3k-8 = -k+4

3k+k = 4+8

4k = 12

k = 3

(2) if 7 times the 7th term of an AP is equal to 11 times the 11th term , show that the 18th term is zero.

Given: 7 times the 7th term of an AP is equal to 11 times the 11th term

7(a+6d) =11(a+10d)

7a+42d=11a+110d

42d-110d=11a-7a

68d = 4a

a =-17d

Now, the 18th term = a+17d=-17d-17d=0

(3) If the nth term of an A.P is 7n-5. Find 100th term

Given that the n th term of the A.P. is 7n-5

So 100 th term will be 7 (100) -5 =695

(4) if m times the mth term of an AP is equal to n times the nth term . Show that (m+n)th term of the AP is zero

We know :- a_n = a +(n-1)d

a _(m+n) = a + (m+n-1)d (just put m+n in place of n ) ------------------------------(1)

Let the first term and common difference of the A.P. be ‘a’ and ‘d’ respectively.

Then, m ^th term = a + (m – 1) d and n ^th term = a + (n – 1) d

By the given condition,

m x a_m = n x a_n

m [a + (m – 1) d] = n [a + (n – 1) d]

⇒ ma + m (m – 1) d = na + n (n – 1) d

=> ma + (m² -m)d - na - (n² -n)d = 0 ( taking the Left Hand Side to the other side )

=> ma -na + (m² - m)d -( n²-n)d = 0 (re-ordering the terms)

=> a (m-n) + d (m²-n²-m+n) = 0 (taking 'a ' and 'd ' common)

=> a (m-n) + d {(m+n)(m-n)-(m-n)} = 0 (a²-b² identity)

Now divide both sides by (m-n)

=> a (1) + d {(m+n)(1)-(1)} = 0

=>a + d (m+n-1) = 0 ---------------(ii)

From equation number 1 and 2 ,

a _(m+n) = a + (m+n-1)d

And we have shown ,

a + d (m+n-1) = 0

So, a _(m+n) = 0 

(5). Prove that the nth term of an AP cannot be n² + 1. Justify your answer.

Common difference of an A.P. must always be a constant.

∴ d cannot be n – 1. Here, d varies when n takes different values.

For n = 1, d = 1 – 1 = 0

For n = 2, d = 2 – 1 = 1

For n = 3, d = 3 – 1 = 2

∴ d is not constant.

Thus, d cannot be taken as n – 1.

a_n is the n ^th term of an A.P. if a_n – a_n –1 = constant

Given, a_n = n ² + 1

a_n – a_n –1 = (n ² + 1) – [(n – 1)² + 1]

= (n ² + 1) – (n ² – 2n + 2)

= 2n – 1        

∴ a_n – a_n –1 ≠ constant

Thus, a_n = n ² + 1 cannot be the n ^th term of A.P.

(6) Find the sum of the first k terms of a series whose n ^th term is 2an+b 

The n ^th term of the AP is given by 2an+b

a1=2a+b

a2 =4a+b

a3=6a+b

Common difference = d=( 4a + b ) - ( 2a + b ) = 2a

Therefore, sum of first k terms =k/2[(2a+(k-1)d]

 = k/2[(2(2a+b)+(k-1)2a]= k/2   x 2 (2a+b+k-a)=k(a+b+ak)

(7) Which term of the AP,  3,10,17 will be 84 more than its 13th term?

Let the nth term be 84 more than the 13^th term.

Now a/q,

a=3, d=10-3=7

So, 13th term= a+12d =3+12x7=87

Then nth term=84+87=171

171=a + (n-1)d

171=3 + (n-1)x7

171-3/7+1=n

168/7+1=n

24+1=25=n

Therefore 25th term of the ap will be 84 more than 13th term

(8) How many terms of the arithmetic series 24 + 21 + 18 + 15 +g, be taken continuously so that their sum is – 351.

In the given arithmetic series, a = 24, d =- 3.

Let us find n such that Sn = – 351

Now, S_n = n/2[(2a + (n-1)d] 

– 351 = n/2[(48 + (n-1)x(-3)] 

on solving we get,  n² - 17n -234 = 0

Þ (n - 26h)(n + 9) = 0

Þ  n = 26 or n = - 9

Here n, being the number of terms needed, cannot be negative

Thus, 26 terms are needed to get the sum -351.

(9) Find the sum of the first 2n terms of the following series. 1² - 2² + 3² - 4² +  

We want to find 1² - 2² + 3² - 4² +.....  to  2n terms

= 1 - 4 + 9 - 16 + 25 ------------2n terms

= (1 – 4) +(9 – 16)+(25 – 36) + ----------- to n terms. (after grouping)

= -3 +(-7)+(-11)+ -------------- n terms

Now, the above series is in an A.P. with first term a = - 3 and common difference d = - 4 

Now, S_n = n/2[(2a + (n-1)d]  = = n/2[(2 x -3) + (n-1)(-4)]  = -n(2n + 1).

(10)  A circle is completely divided into n sectors in such a way that the angles of the sectors are in arithmetic progression. If the smallest-of these angles is 8° and the largest 72°, calculate n and the angle in the fourth sector.

Let the common difference of the A.P. be x

The smallest angle = 8°

⇒ a = 8

And the largest is 72°

⇒ a_n = 72

⇒ a + (n – 1)d = 72

⇒ 8 + (n – 1)d = 72

⇒ (n – 1) d = 72 – 8 = 64 ... (1)

We know that sum of all the angles of a circle is 360°

S_n = n/2[(2a + (n-1)d]  = 360

Þ S_n = n/2[(2x8 + 64] = 360

Þ n= 9

Putting the value of n in equation (1) we get

(9 – 1) d = 64

d = 8

Now angle in fourth sector = a₄ = a + (4 – 1) d

= a + 3d = 8 + 3 × 8 = 8 + 24 = 32

∴ The value of n = 9 and angle in fourth sector is 32°

10th Maths SA-2 Chapter Links

Quadratic Equations

Arithmetic Progressions

Height and Distance

Co-ordinate Geometry

Circles

Area Related to Circles

Surface Areas and Volumes

Probability