If 19^ th terms of non -zero A.P. is zero, then its (49^ th term)… - Vidyadip

MCQ

If $19^{th}$ terms of non -zero $A.P.$ is zero, then its ($49^{th}$ term) : ($29^{th}$ term) is

A
$4 : 1$
B
$1 : 3$
✓
$3 : 1$
D
$2 : 1$

✓

Answer

Correct option: C.

$3 : 1$

c
$a + 18d = 0 \Rightarrow a = - 18d$

$\frac{{{t_{49}}}}{{{t_{29}}}} = \frac{{a + 48d}}{{a + 28d}} = \frac{{ - 18d + 48d}}{{ - 18d + 28d}}$

$ = \frac{{30d}}{{10d}} = 3$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The function $\mathrm{f}(\mathrm{x})$, that satisfies the condition $\mathrm{f}(\mathrm{x})=\mathrm{x}+\int_{0}^{\pi / 2} \sin \mathrm{x} \cdot \cos y \mathrm{f}(\mathrm{y}) \mathrm{dy}$, is :

Let $f(\mathrm{x})=\left(\sin \left(\tan ^{-1} \mathrm{x}\right)+\sin \left(\cot ^{-1} \mathrm{x}\right)\right)^{2}-1,|\mathrm{x}|>1$ If $\frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}\left(\sin ^{-1}(f(x))\right) $ and $ y(\sqrt{3})=\frac{\pi}{6}$ then $y(-\sqrt{3})$ is equal to

$\int_{}^{} {\frac{1}{{x - {x^3}}}\;dx = } $

The centres of the circles ${x^2} + {y^2} = 1$, ${x^2} + {y^2} + 6x - 2y = 1$ and ${x^2} + {y^2} - 12x + 4y = 1$ are

The value of the integral $\int\limits_{ - 2}^2 {\frac{{{{\sin }^2}\,x}}{{\left[ {\frac{x}{\pi }} \right] + \frac{1}{2}}}\,\,dx} $ (where $[x]$ denotes the greatest integer less than or equal to $x$ ) is

Suppose that $20$ pillars of the same height have been crected along the boundary of a circular stadium. If the top of each pillar has been connected by beams with the top of all its non-adjacent pillars, then the total number of beams is

$7\log \left( {{{16} \over {15}}} \right) + 5\log \left( {{{25} \over {24}}} \right) + 3\log \left( {{{81} \over {80}}} \right)$ is equal to

The values of $'a'$ and $'b'$ for which equation ${x^4} - 4{x^3} + a{x^2} + bx + 1 = 0$ have four real roots

A ray of light along $x + \sqrt 3 y = \sqrt 3 $ gets reflected upon reaching $x- $ axis , the equation of the reflected ray is

Let $f'(x) > 0$ and $g'(x) < 0$ for all $x \in R$ then