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 m2: n2. 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,
a = 8, b = 26, n =7
So,
26 = 8 + (7 – 1) d
6d = 26 –8 = 18
d = 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 +4d = 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.