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32.  150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on.

Let x be the number of days in which 150 workers finish the work. According to the given information, we have  Series   is a AP first term=a=150 common difference= -4 number of terms = x+8 Since x cannot be negative so x=17. Thus, in 17 days 150 workers finish the work. Thus, the required number of days = 17+8=25 days.

31. A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.

Cost of machine = Rs. 15625 Machine depreciate each year by 20%. Therefore, its value every year is 80% of the original cost i.e.  of the original cost.  Value at the end of 5 years                                                                                             Thus, the value of the machine at the end of 5 years is Rs. 5120

30.  A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.

Given : A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.               Interest in fifteen year 10000+ 14 times Rs. 500  Amount in 15 th year                                                                         Amount in 20 th year

29.  A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.

The numbers of letters mailed forms a GP :  first term  = a=4 common ratio=r=4 number of terms = 8 We know that the sum of GP is                                                                                                                                                                                                                         costs to mail one letter are 50 paise. Cost of...

28.  Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

Given: Shamshad Ali buys a scooter for Rs 22000. Therefore , unpaid amount = 22000-4000=Rs. 18000 According to the given condition, interest paid annually is 10% of 18000,10% of 17000,10% of 16000,......................10% of 1000. Thus, total interest to be paid                                                                                                                   Here,  is a AP with...

27.   A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?

Given : Farmer pays Rs 6000 cash. Therefore , unpaid amount = 12000-6000=Rs. 6000 According to given condition, interest paid annually is 12% of 6000,12% of 5500,12% of 5000,......................12% of 500. Thus, total interest to be paid                                                                                                                 Here,  is a AP with first term =a=500 and...

26. Show that $\frac{1 \times 2 ^ 2 + 2 \times 3^2 + ... + n \times (n+1)^2}{1^2 \times 2 + 2 ^2 \times 3 + ...+ n^2 \times ( n+1)} = \frac{3n+5}{3n+1}$

To prove :                     the nth term of numerator  nth term of the denominator                      Numerator :                                                                                                                                                                                                             Denominator :                                                        ...

25.   Find the sum of the following series up to n terms: $\frac{1^3}{1} + \frac{1^3+2^3}{1+3}+ \frac{1^3+2^3+3^3}{1+3+5}+ ...$

n term of series :                                                                                                                    Here,  are in AP with first term =a=1 , last term = 2n-1, number of terms =n                                                                                                                                                                                      ...

24.  If $S_1 , S_2 , S_3$ are the sum of first n natural numbers, their squares and their cubes, respectively, show that $9 S ^2 _2 = S_3 ( 1+ 8 S_1)$

To prove :  From given information,                          Here ,                                                                                                                    Also,                                  From equation 1 and 2 , we have           Hence proved .

23.   Find the sum of the first n terms of the series: 3+ 7 +13 +21 +31 +…

The series: 3+ 7 +13 +21 +31 +….............. n th term  =

22.  Find the 20th term of the series $2 \times 4+4\times 6+\times 6\times 8+....+n$ terms.

the series =                                      Thus, the 20th term of series is 1680

21(ii)   Find the sum of the following series up to n terms:

.6 +. 66 +. 666+…

Sum of  0.6 +0. 66 + 0. 666+…................. It can be written  as   to n terms

21(i)  Find the sum of the following series up to n terms:

$5 + 55+ 555 + ....$

is not a GP. It can be changed in GP by writing terms as   to n terms Thus, the sum is

20.   If a, b, c are in A.P.; b, c, d are in G.P. and 1/c , 1/d , 1/e are in A.P. prove that a, c, e are in G.P.

Given: a, b, c are in A.P Also,  b, c, d are in G.P. Also, 1/c, 1/d, 1/e are in A.P   To prove: a, c, e are in G.P. i.e. From 1, we get                            From 2, we get                              Putting values of b and d, we get Thus, a, c, e are in G.P.

19.  The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that $a: b = \left ( m + \sqrt {m^2 -n^2 }\right ) : \left ( m- \sqrt {m^2 - n^2} \right )$
.

Let two numbers be a and b. According to the given condition,      ...................................................................1 We get,                           .....................................................2 From 1 and 2, we get Putting the value of a in equation 1, we have

18.  If a and b are the roots of  $x^2 -3 x + p = 0$ and c, d are roots of  $x^2 -12 x + q = 0$, where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17:15.

Given: a and b are the roots of   Then,  Also, c, d are roots of   Given: a, b, c, d form a G.P Let,  From 1 and 2, we get                        and                                         On dividing them,           When , r=2 ,                           When , r=-2,                      CASE (1) when r=2 and x=1, i.e. (q + p) : (q – p) = 17:15. CASE (2) when r=-2 and x=-3, i.e. (q +...

17. If a, b, c, d are in G.P, prove that $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ are in G.P.

Given: a, b, c, d are in G.P. To prove: are in G.P. Then we can write,                  Let  be in GP LHS:                                                                                                                   Hence  proved  Thus, are in GP

16.  If  $a (\frac{1}{b}+\frac{1}{c}) , b ( \frac{1}{c}+\frac{1}{a}) , c ( \frac{1}{a}+ \frac{1}{b})$  are in A.P., prove that a, b, c are in A.P.

Given: are in A.P.   Thus, a,b,c  are in AP.

15.  The pth, qth and rth terms of an A.P. are a, b, c, respectively. Show that

$(q - r )a + (r - p )b + (p - q )c = 0$

Given: The pth, qth and rth terms of an A.P. are a, b, c, respectively. To prove :   Let the first term of AP be 't' and common difference be d Subtracting equation 2 from 1, we get Subtracting equation 3 from 2, we get Equating values of d, from equation 4 and 5, we have Hence proved.

14.  Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that $P^2 R^n = S ^n$

Ler there be a GP  According to given information, To prove :  LHS :                                                                                       Hence proved
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