Question
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.
✓
Answer
The given terms can be written as follows:
$1, 9, 25, 49,....$
Here, $a_2- a_1= 9 - 1 = 8$
$a_3- a_2= 25 - 9 = 16$
$a_4- a_3= 49 - 25 = 24$
Since $a_{k+1}- a_k$ is not same for all values of k
Hence, it is not an AP. So we can not find next three terms.
$1, 9, 25, 49,....$
Here, $a_2- a_1= 9 - 1 = 8$
$a_3- a_2= 25 - 9 = 16$
$a_4- a_3= 49 - 25 = 24$
Since $a_{k+1}- a_k$ is not same for all values of k
Hence, it is not an AP. So we can not find next three terms.
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
Prove the trigonometric identity:
$
\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{\sin ^2 A-\cos ^2 A}=\frac{2}{2 \sin ^2 A-1}=\frac{2}{1-2 \cos ^2 A}
$
→$
\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{\sin ^2 A-\cos ^2 A}=\frac{2}{2 \sin ^2 A-1}=\frac{2}{1-2 \cos ^2 A}
$
Prove the following trigonometric identities.
$\tan^2\text{A}+\cot^2\text{A}=\sec^2\text{A}\text{ cosec}^2\text{A}-2$
→$\tan^2\text{A}+\cot^2\text{A}=\sec^2\text{A}\text{ cosec}^2\text{A}-2$
$(B)$ The $\text{H.C.F}$ of $85$ and $238$ is expressible in the form $85m -238.$ Find the value of $m.$
→The circumforence of a circle is 39.6cm. Find its area. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
→Observe the Fig. given below and then find $\angle$P.
→Write the median class of the following distribution:
|
Class
|
0-10
|
10-20
|
20-30
|
30-40
|
40-50
|
50-60
|
60-70
|
|
Frequency
|
4
|
4
|
8
|
10
|
12
|
8
|
4
|
Show that the number $5 \times 11 \times 17+3 \times 11$ is a composite number.
→The graph of the polynomial $f(x)=a x^2+b x+c$ is as shown in Fig. Write the value of $b^2-4 a c$ and the number of real zeros of $f(x)$, write the sign of $c$.
→In the given figure, $\text{ABCD}$ is a quadrilateral. Diagonal $BD$ bisects $\angle B$ and $\angle D$ both. Prove that :
$(i) \triangle \text{ABD} \sim \triangle CBD$
$(ii) \text{AB = BC}$
→$(i) \triangle \text{ABD} \sim \triangle CBD$
$(ii) \text{AB = BC}$
Very-Short-Answer Questions.
A die is thrown once. Find the probability of getting:
A number between 3 and 6.
A die is thrown once. Find the probability of getting:
A number between 3 and 6.