Question
Find where 0 (zero) is a term of the A.P. $40,37,34,31, \ldots \ldots$.
✓
Answer
$A.P. 40, 37, 34, 31, .....$
Here,
$a = 40, d = -3$
Let $T_n = 0$
$T_n = a + (n - 1)d$
$\Rightarrow 0 = 40 + (n - 1)(-3)$
$\Rightarrow 0 = 40 - 3n + 3$
$\Rightarrow 3n = 43$
$\Rightarrow\ \text{n}=\frac{43}{3}$ which is in fraction
There is no term which is $0$.
Here,
$a = 40, d = -3$
Let $T_n = 0$
$T_n = a + (n - 1)d$
$\Rightarrow 0 = 40 + (n - 1)(-3)$
$\Rightarrow 0 = 40 - 3n + 3$
$\Rightarrow 3n = 43$
$\Rightarrow\ \text{n}=\frac{43}{3}$ which is in fraction
There is no term which is $0$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Is the given sequence: $1^2, 3^2, 5^2, 7^2, .....$. forms an $AP$? If it forms an AP, then find the common difference d and write the next three terms.
→In an AP: l = 28, S = 144, and there are total 9 terms. Find 'a'.
Are the given numbers $1, 3, 9, 27, ........$ forms an $AP$? If It forms an $AP$, then find the common difference d and write the next three terms.
→In the given figure, $\triangle\text{AHK}$ is similar to $\triangle\text{ABC}.$ If AK = 10cm, BC = 3.5cm and HK = 7cm, find AC.
→Determine the nature of the root of the following quadratic equation:
$3\text{x}^2-2\sqrt{6}\text{x}+2=0$
→$3\text{x}^2-2\sqrt{6}\text{x}+2=0$
In Figure, OACB is a quadrant of a circle with centre O and radius 7 cm . If $OD =3 cm$, then find the area of the shaded region.
→Is $0.2$ a root of the equation $x^2 – 0.4 = 0$? Justify.
→If $\sin\theta=\frac{1}{3},$ then find is the value of $2\cot^2\theta+2$.
→In $\triangle ABC , D$ and $E$ are the points on the sides AB and AC respectively such that DE||BC. If AD = 6x - 7, DB = 4x - 3, AE = 3x - 3 and EC = 2x - 1, find the value of x.
→For what value of k are the roots of the quadratic equation $\text{kx}\big(\text{x}-2\sqrt5\big)+10=0$ real and equal?
→