A manufacturer of TV sets produced 600 sets i | vedclass

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

Question

(D) Since the production increases uniformly by a fixed number every year,the number of $TV$ sets manufactured in the $1^{st}, 2^{nd}, 3^{rd}, \dots$

Vedclass · Accepted Answer

$550$

Vedclass · Answer

(D) Since the production increases uniformly by a fixed number every year,the number of $TV$ sets manufactured in the $1^{st}, 2^{nd}, 3^{rd}, \dots$ years forms an Arithmetic Progression $(AP)$.Let $a$ be the production in the $1^{st}$ year and $d$ be the uniform increase in production every year.The production in the $n^{th}$ year is given by $a_n = a + (n-1)d$.Given,$a_3 = 600$ and $a_7 = 700$.Thus,$a + 2d = 600$ (Equation $1$) and $a + 6d = 700$ (Equation $2$).Subtracting Equation $1$ from Equation $2$: $(a + 6d) - (a + 2d) = 700 - 600$.$4d = 100$,which gives $d = 25$.Substituting $d = 25$ in Equation $1$: $a + 2(25) = 600$.$a + 50 = 600$,so $a = 550$.Therefore,the production in the $1^{st}$ year is $550$ sets.

$A$ manufacturer of $TV$ sets produced $600$ sets in the third year and $700$ sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year,find the production in the $1^{st}$ year.

Similar Questions

Ramkali saved ₹ $5$ in the first week of a year and then increased her weekly savings by ₹ $1.75$. If in the $n$-th week,her weekly savings become ₹ $20.75$,find $n$.

How many three-digit numbers are divisible by $7$?

Which term of the $AP: 21, 18, 15, \ldots$ is $-81$? Also,is any term $0$? Give reason for your answer.

In the following arithmetic progression $(AP)$,find the missing terms in the boxes: $\square, 13, \square, 3$.

Find the sum of first $22$ terms of an $AP$ in which $d=7$ and $22^{nd}$ term is $149$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module