Advertisements
Advertisements
प्रश्न
Find the sum of the following arithmetic series:
`7 + 10 1/2 + 14 + ....... + 84`
Advertisements
उत्तर
The given arithmetic series is `7 + 10 1/2 + 14 + ....... + 84`
` Here , a = 7 , d = 10 1/2 - 7 = 21/ 2 - 7 = (21-4)/2 = 7/2 and l = 84.`
Let the given series contains n terms. Then,
an = 84
`⇒ 7 + (n-1) xx 7/2 = 84 [ a_n = a + (n-1) d ]`
` ⇒ 7/2 n + 7/2 = 84`
`⇒ 7/2 n = 84 - 7/2 = 161/2`
`⇒ n = 161/7 = 23`
∴ Required sum`= 23/2 xx (7 + 84) [ s_n = n/2 (a+l) ]`
`=23/2 xx 91`
=`2030/2`
`1046 1/2`
APPEARS IN
संबंधित प्रश्न
Find the term t15 of an A.P. : 4, 9, 14, …………..
Find the sum of the following arithmetic series:
34 + 32 + 30 +...+10
Decide whether the following sequence is an A.P., if so find the 20th term of the progression:
–12, –5, 2, 9, 16, 23, 30, ..............
Find the 27th term of the following A.P.
9, 4, –1, –6, –11,...
Six year before, the age of mother was equal to the square of her son's age. Three year hence, her age will be thrice the age of her son then. Find the present ages of the mother and son.
Select the correct alternative and write it.
What is the sum of first n natural numbers ?
Select the correct alternative and write it.
If a share is at premium, then -
For a given A.P. a = 3.5, d = 0, then tn = _______.
Find the 23rd term of the following A.P.: 9, 4,-1,-6,-11.
Choose the correct alternative answer for the following sub-question
If the third term and fifth term of an A.P. are 13 and 25 respectively, find its 7th term
How many two-digit numbers are divisible by 5?
Activity :- Two-digit numbers divisible by 5 are, 10, 15, 20, ......, 95.
Here, d = 5, therefore this sequence is an A.P.
Here, a = 10, d = 5, tn = 95, n = ?
tn = a + (n − 1) `square`
`square` = 10 + (n – 1) × 5
`square` = (n – 1) × 5
`square` = (n – 1)
Therefore n = `square`
There are `square` two-digit numbers divisible by 5
Decide whether 301 is term of given sequence 5, 11, 17, 23, .....
Activity :- Here, d = `square`, therefore this sequence is an A.P.
a = 5, d = `square`
Let nth term of this A.P. be 301
tn = a + (n – 1) `square`
301 = 5 + (n – 1) × `square`
301 = 6n – 1
n = `302/6` = `square`
But n is not positive integer.
Therefore, 301 is `square` the term of sequence 5, 11, 17, 23, ......
12, 16, 20, 24, ...... Find 25th term of this A.P.
If tn = 2n – 5 is the nth term of an A.P., then find its first five terms
In an A.P. if the sum of third and seventh term is zero. Find its 5th term.
Write the next two terms of the A.P.: `sqrt(27), sqrt(48), sqrt(75)`......
