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Find the 20th term of the given AP 5, 8, 11, …
Ajay started work in 1995 at an annual salary of Rs5000 and received an increment of Rs200 each year. In which year did his income reach of Rs7000.
Rs7000−Rs5000=Rs2000 has increased by increasing every year Rs200.
years, i.e., in the 11th year he will receive Rs7000.
In an A.P., the sum of first n terms in\(\dfrac{{3{n^2}}}{2} + \dfrac{{5n}}{2}\). find its 25th term.
If sum of n terms in an A.P. is given\({S_n} = 2{n^2} + 2\), then find the 20th term.
If a, b, c are in A.P. and ac=b then find the value of a in terms of b.
a−c=b
(a, b, c are in AP)
From equation (1),
Find the number of terms in A.P. if sum of these terms is 710 and last terms are 7 and 64 respectively.
Sum =70=(7+64)k/2
If first, second and last terms of an A.P. are 6, 26, 30 respectively then the number of terms are
Solving equations (1) and (2)
n=10
Which term of the sequence 17, 25, 33, 41 … is just greater than 90.
Let
If a, A1, A2, b are in A.P. then find\(\dfrac{{{A_1}}}{{{A_2}}}\)
and n=2
If the product of roots of equation\(\left( {a – b} \right){x^2} + \left( {b – c} \right)x + \left( {c – a} \right) = 0\)is 1, then:
Find the number of terms in 7, 13, 19, …, 205
a=7, b=6
Find the 31st term of an AP whose 11th term is 38 and 16th term is 73
Subtracting equation (1) from (2), we get
5d=35
d=7
If the 3rd and 9th terms of an AP are 4 and 8 respectively. Which term of this AP is zero?
How many 3 digit numbers are divisible by 7?
Smallest 3 digit number divisible by 7 is 105.
Greatest 3 digit number divisible by 7 is 994.
Number of terms=(last termfirst term)/common difference+1
For what value of n are the nth terms of two APs: 63, 65, 67, … and 3, 10, 17,… equal?
In the first AP a=63, d=2
In the second AP a=3 and d=7
According to question,
As the result is not an integer so there won’t be any term with equal values for both APs.
\(\left( 1 \right) + \left( {1 + 1} \right) + \left( {1 + 1 + 1} \right) + … + \left( {1 + 1 + 1 + … + n – 1times} \right) = \)
Which term of the AP: 24, 21, 18 … is the first negative term?
Let
10th term will be negative.
If the 25th element of an AP is 352 and the difference between the 48th element and the 38th element is 140. What is the first element of the Ap?
On solving equation (1) and (2), a=16.
Neha required Rs4440 after 12 weeks to buy a cell phone. She saved Rs120 in the first week ad increased her weekly savings by Rs20 every week. Find whether she will able to but a cell phone after 12 weeks.
First term is 120 and common difference is 20. She has to buy a cell phone after 12 weeks. So she will save the amount for 12 weeks. Thus the total number of term in the AP will be 12.
If the sum of first n, 2n, 3n terms of an AP is S1, S2, S3 respectively then S3 is equal to
S1=sum of first n terms
S2=sum of first 2n terms
S3=sum of first 3n terms
On solving,
If the sum of the first p terms of an AP be q and the sum of first q terms of an AP be p then the sum of its first (p+q) terms is
sp=q
The sum of first (p+q) terms
=−(p+q)
Which is the 9th term from the end of the arithmetic progression 11, 7, 25…, 745?
How many elements are these in AP 11, 3, 5…, 301?
What is the sum of two digit odd positive numbers?
Smallest 2 digit number=11
Greatest 2 digit number =99
Number of terms
If the mean of x and 1/x is M, then the mean of x3 and 1/x3 is
The AM of\({\left( {a – b} \right)^3}\)is
Mean
There are three arithmetic means between 2 and 12 the mean are
If the AM of a and b is\(\dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}\), then the value of n is
Sum of n terms of the series\(\left( {1 – \dfrac{1}{n}} \right) + \left( {1 – \dfrac{2}{n}} \right) + \left( {1 – \dfrac{3}{n}} \right)…\)upto n terms
(1+1+…n terms)−(1/n+2/n+…n terms)
The interior angles of a polygon are in AP. If the smallest angle be 120 and the common difference be 50. Find the number of sides.
a=120, d=5
sum of interior angles =
sum of n terms
On solving, n=9 or 16
Which term of AP 5, 15, 25… will be 130 more than its 31st term?
The digits of a positive integer, having three digits are in AP and their sum is 15. The number obtained by reversing the digits is 594mless than the original number. Find the number.
Let the digits at ones, tens and hundreds place be (ad), a, (a+d)
Number is
Reversed number=111a99d
Sum of the digits=:
Number=111×5+99×3=852
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 logs in the next row, 18 in the row next to it and so on. In how many rows 200 logs are placed?
If n=25, number of logs, which is not meaningful, so, n=16
150 workers were engaged to finish a place of work in certain number of days. Four workers dropped the second day, four more workers dropped the third day and so on. It takes 8 days to finish the work now. Find the number of days in which the work was completed.
If the sum of m terms of an AP is same as the sum of its n terms. The sum of (m+n) term in