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S seema garhwal
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.

S seema garhwal
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

S seema garhwal
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

S seema garhwal
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...

S seema garhwal
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...

S seema garhwal
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...

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

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

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

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

S seema garhwal
the series =                                      Thus, the 20th term of series is 1680

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

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

S seema garhwal
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.

S seema garhwal
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

S seema garhwal
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 +...

S seema garhwal
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

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