Advertisements
Advertisements
प्रश्न
Write an A.P. whose first term is a and common difference is d in the following.
a = 6, d = –3
Advertisements
उत्तर
a = 6, d = –3
t1 = a = 6 ...[Given]
t2 = t1 + d = 6 + (-3) = 6 - 3 = 3
t3 = t2 + d = 3 + (-3) = 3 - 3 = 0
t4 = t3 + d = 0 + (-3) = 0 - 3 = -3
∴ The required A.P. is 6, 3, 0, -3, ...
APPEARS IN
संबंधित प्रश्न
Ramkali required Rs 2,500 after 12 weeks to send her daughter to school. She saved Rs 100 in the first week and increased her weekly saving by Rs 20 every week. Find whether she will be able to send her daughter to school after 12 weeks.
What value is generated in the above situation?
Find the sum of first 40 positive integers divisible by 6.
Find the sum of all odd numbers between 100 and 200.
Find the sum of all integers between 100 and 550, which are divisible by 9.
Find the sum of the first 11 terms of the A.P : 2, 6, 10, 14, ...
Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.
How many terms of the AP 63, 60, 57, 54, ... must be taken so that their sum is 693? Explain the double answer.
The fourth term of an A.P. is 11. The sum of the fifth and seventh terms of the A.P. is 34. Find its common difference.
Choose the correct alternative answer for the following question .
First four terms of an A.P. are ....., whose first term is –2 and common difference is –2.
If first term of an A.P. is a, second term is b and last term is c, then show that sum of all terms is \[\frac{\left( a + c \right) \left( b + c - 2a \right)}{2\left( b - a \right)}\].
If `4/5` , a, 2 are three consecutive terms of an A.P., then find the value of a.
If Sr denotes the sum of the first r terms of an A.P. Then , S3n: (S2n − Sn) is
An article can be bought by paying Rs. 28,000 at once or by making 12 monthly installments. If the first installment paid is Rs. 3,000 and every other installment is Rs. 100 less than the previous one, find:
- amount of installments paid in the 9th month.
- total amount paid in the installment scheme.
How many terms of the A.P. 24, 21, 18, … must be taken so that the sum is 78? Explain the double answer.
Determine the sum of first 100 terms of given A.P. 12, 14, 16, 18, 20,......
Activity :- Here, a = 12, d = `square`, n = 100, S100 = ?
Sn = `"n"/2 [square + ("n" - 1)"d"]`
S100 = `square/2 [24 + (100 - 1)"d"]`
= `50(24 + square)`
= `square`
= `square`
If the first term of an AP is –5 and the common difference is 2, then the sum of the first 6 terms is ______.
Find the value of a25 – a15 for the AP: 6, 9, 12, 15, ………..
Complete the following activity to find the 19th term of an A.P. 7, 13, 19, 25, ........ :
Activity:
Given A.P. : 7, 13, 19, 25, ..........
Here first term a = 7; t19 = ?
tn + a + `(square)`d .........(formula)
∴ t19 = 7 + (19 – 1) `square`
∴ t19 = 7 + `square`
∴ t19 = `square`
Measures of angles of a triangle are in A.P. The measure of smallest angle is five times of common difference. Find the measures of all angles of a triangle. (Assume the measures of angles as a, a + d, a + 2d)
The nth term of an A.P. is 6n + 4. The sum of its first 2 terms is ______.
