Different A.P.'s are constructed with the | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$1^{	ext {st }} 	ext { term }=100=a$

Last term $=199=\ell$

If $3$ term

$a, a+d, a+2 d$

$a _{ a }=\ell= a +( n -1) d$

$d _{ i }=\frac{

Vedclass · Accepted Answer

$53$

Vedclass · Answer

$1^{	ext {st }} 	ext { term }=100=a$

Last term $=199=\ell$

If $3$ term

$a, a+d, a+2 d$

$a _{ a }=\ell= a +( n -1) d$

$d _{ i }=\frac{\ell- a }{ n - l }$

$n ightarrow$ number of terms

$n =3, d _{1}=\frac{199-100}{2}$

$=\frac{99}{2} 
otin I$

$n =4, d _{2}=\frac{99}{3}=33 \in I$

$n =10, d _{3}=\frac{99}{9}=11 \in I$

$n =12, d _{4}=\frac{99}{11}=9 \in I$

$	herefore \sum d _{ i }=33+11+9=53$

Different $A.P.$'s are constructed with the first term $100$,the last term $199$,And integral common differences. The sum of the common differences of all such, $A.P$'s having at least $3$ terms and at most $33$ terms is.

Similar Questions

The mean of the series $a,a + nd,\,\,a + 2nd$ is

The first term of an $A.P.$ of consecutive integers is ${p^2} + 1$ The sum of $(2p + 1)$ terms of this series can be expressed as

If ${a_1} = {a_2} = 2,\;{a_n} = {a_{n - 1}} - 1\;(n > 2)$, then ${a_5}$ is

Let ${S_1},{S_2},......,{S_{101}}$ be the consecutive terms of an $A.P$ . If $\frac{1}{{{S_1}{S_2}}} + \frac{1}{{{S_2}{S_3}}} + .... + \frac{1}{{{S_{100}}{S_{101}}}} = \frac{1}{6}$ and ${S_1} + {S_{101}} = 50$ , then $\left| {{S_1} - {S_{101}}} \right|$ is equal to

Find the sum of all natural numbers lying between $100$ and $1000,$ which are multiples of $5 .$