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Chapter 9 Sequences and Series Ex 9.2 Solutions

Question - 11 : -

Sum of the first p, q and r terms of anA.P. are a, b and c, respectively.

Prove that  

Answer - 11 : -

Let a1 and d bethe first term and the common difference of the A.P. respectively.

Then according to thequestion, we have

Now, subtracting (2)from (1), we get

Question - 12 : -

The ratio of the sums of m and n termsof an A.P. is m2n2. Show thatthe ratio of mth and nth termis (2m – 1): (2n – 1).

Answer - 12 : -

Let’s considerthat a and b to be the first term and thecommon difference of the A.P. respectively.

Then from thequestion, we have

Hence, the givenresult is proved.

Question - 13 : -

If the sum of n terms of an A.P. is 3n2 +5n and its mth term is 164, find the value of m.

Answer - 13 : -

Let’s consider a and b tobe the first term and the common difference of the A.P. respectively.

am = a + (m –1)d = 164 … (1)

We the sum of theterms is given by,

Sn =n/2 [2a + (n-1)d]

Question - 14 : - Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Answer - 14 : -

Let’s assume A1,A2, A3, A4, and A5 to be fivenumbers between 8 and 26 such that 8, A1, A2, A3,A4, A5, 26 are in an A.P.

Here we have,

= 8, = 26, n =7

So,

26 = 8 + (7 – 1) d

6d = 26 –8 = 18

= 3

Now,

A1 = a + d =8 + 3 = 11

A2 = a +2d = 8 + 2 × 3 = 8 + 6 = 14

A3 = a +3d = 8 + 3 × 3 = 8 + 9 = 17

A4 = a +4= 8 + 4 × 3 = 8 + 12 = 20

A5 = a +5d = 8 + 5 × 3 = 8 + 15 = 23

Therefore, therequired five numbers between 8 and 26 are 11, 14, 17, 20, and 23.

Question - 15 : -

If is the A.M. between a and b,then find the value of n.

Answer - 15 : -

The A.M between a andb is given by, (a + b)/2

Then according to thequestion,

Thus, the value of nis 1.

Question - 16 : -

Between 1 and 31, m numbers have been inserted in such away that the resulting sequence is an A.P. and the ratio of 7th and(m – 1)th numbers is 5: 9. Find the value of m.

Answer - 16 : -

Let’s consider a1,a2, … am be m numbers suchthat 1, a1, a2, … am, 31 is an A.P.

And here,

a = 1, b = 31, n = m +2

So, 31 = 1 + (m +2 – 1) (d)

30 = (m +1) d

d = 30/ (m + 1) …….(1)

Now,

a1 = a + d

a2 = a +2d

a3 = a +3d …

Hence, a7 = a +7d

am–1 = a +(m – 1) d

According to thequestion, we have

Therefore, the valueof m is 14.

Question - 17 : -

A man starts repaying a loan as first instalment of Rs. 100. If heincreases the instalment by Rs 5 every month, what amount he will pay in the 30th instalment?

Answer - 17 : -

Given,

The first instalmentof the loan is Rs 100.

The second instalmentof the loan is Rs 105 and so on as the instalment increases by Rs 5 everymonth.

Thus, the amount thatthe man repays every month forms an A.P.

And the, A.P. is 100,105, 110, …

Where, firstterm, a = 100

Commondifference, d = 5

So, the 30th termin this A.P. will be

A30 = a + (30 – 1)d

= 100 + (29) (5)

= 100 + 145

= 245

Therefore, the amountto be paid in the 30th instalment will be Rs 245.

Question - 18 : -

The difference between any two consecutive interior angles of a polygonis 5°. If the smallest angle is 120°, find the number of the sides of thepolygon.

Answer - 18 : -

It’s understood fromthe question that, the angles of the polygon will form an A.P. with commondifference d = 5° and first term a = 120°.

And, we know that thesum of all angles of a polygon with n sides is 180° (n –2).

Thus, we can say

Thus, a polygon having9 and 16 sides will satisfy the condition in the question.

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