Question

PROGRESSIONS
4. How many terms of the AP:9, 17, 25, ... must be taken to give a sum of 636?
5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms
and the common difference.
6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference
is 9. how many terms are there and what is their sum?
7. Find the sum of first 22 terms of an AP in which d=7 and 22nd term is 149.
8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18
respectively.
9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of
first n terms.
10. Show that a,, 22, ...,a, ... form an AP where a is defined as below:
(1) a = 3 +4n
(ii) a = 9 - 50
Also find the sum of the first 15 terms in each case.
11. If the sum of the first n terms of an AP is 4n-n?, what is the first term (that is S.)? Wha
of first two terms? What is the second term? Similarly, find the 3rd, the 10th and​

Answer 1

Step-by-step explanation:

all next question are also simple

just use NCERT formula to calculate them

mark me BRAINLIEST

Answer 2

Answer:

solution of q 4

Step-by-step explanation:

AP. 9,17,25

a=9

d=a2-a1

=17-9

=8

Sn=636

n=?

Sn =n/2[2a+(n-1)d]

636=n/2[2(9)+(n-1)(8)]

636=n/2[18+8n-8]

636=n/2[18-8+8n]

636=n/2[10+8n]

here in RHS.. I took 2 as common from (10+8n)

636=n/2*2(5+4n)

so here in RHS 2 gets cancelled in n/2 *2

we get ,..

636=n(5+4n)

636=5n+4n^2

4n^2+5n-636=0

4n^2+53n-48n-636=0

n(4n+53)-12(4n+53)=0

(n-12)(4n+53)=0

n-12=0, 4n+53=0

n=12 , 4n= -53

n= 12, n= -53/4

Therefore n=12 ( since "n" cannot b negative or in fraction)

I hope u will surely understand my solution