Given an A.P. whose terms are all positive in | vedclass

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

Question

(C) Let $a$ be the first term and $d$ be the common difference of the given $A.P.$The second term is $a + d = 12$ .....$(1)$The sum of the first nine

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(C) Let $a$ be the first term and $d$ be the common difference of the given $A.P.$The second term is $a + d = 12$ .....$(1)$The sum of the first nine terms is given by:${S_9} = \frac{9}{2}(2a + 8d) = 9(a + 4d)$Given that $200 $200 Substitute $a = 12 - d$ from equation $(1)$ into the inequality:$200 $200 $200 Subtract $108$ from all parts:$92 Since the terms are positive integers,$d$ must be an integer. Testing values for $d$:If $d = 3$,$27 	imes 3 = 81$ (too small).If $d = 4$,$27 	imes 4 = 108$ (which satisfies $92 If $d = 5$,$27 	imes 5 = 135$ (too large).Thus,$d = 4$.From equation $(1)$,$a + 4 = 12$,so $a = 8$.The $4^{th}$ term is $a + 3d = 8 + 3(4) = 8 + 12 = 20$.

Given an $A.P.$ whose terms are all positive integers. The sum of its first nine terms is greater than $200$ and less than $220$. If the second term is $12$,then its $4^{th}$ term is:

Similar Questions

If the sides of a right-angled triangle are in $A.P.$,then their ratio is:

Let $x, y > 0$. If $x^{3} y^{2} = 2^{15}$,then the least value of $3x + 2y$ is

The $15^{th}$ term of the arithmetic progression $4 + 9 + 14 + 19 + \dots$ is......

If $p$ times the $p^{th}$ term of an $A.P.$ is equal to $q$ times the $q^{th}$ term of an $A.P.$,then the $(p + q)^{th}$ term is

The sums of $n$ terms of two arithmetic series are in the ratio $(2n + 3) : (6n + 5)$. Then the ratio of their $13^{th}$ terms is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module