1 + 4 + 9 + 16 + ... + 225
= 1²+ 2² + 3²  + ... + n²  \(=\frac{n(n+1)(2 n+1)}{6}\)
1 + 4 + 9 + 16 + ... + 225  = 1² + 2² + 3² + ... + 15²
\(=\frac{15(15+1)[2(15)+1]}{6} \)
\(=\frac{15 \times 16 \times 31}{6}=1240 \)
1 + 4+ 9 + 16 + ... + 225 = 1240

9,13,17, 21, 25, ... 
t₂ - t₁ = 13 - 9 = 4
t₃ - t₂ = 17 - 13 = 4
t₄ - t₃ = 21 - 17 = 4
t₅ - t₄ = 25 - 21 = 4
t₂ - t₁ = t₃ - t₂ =  t₄ - t₃ = t₅ - t₄ = 4
The difference between consecutive terms are equal.

98 \(\equiv \) (x + 4) (mod 5)
98 - (x + 4) = 5n , for some integer n.
94 - x  = 5n
94- x is a multiple of 5
Therefore , the least positive value of x must be 4
Since 94 - 4 = 90  is the nearest multiple of 5 less than 94.

1³ + 2³ + 3³ + ...+ 16³ = \(\left[ \frac { 16\times \left( 16+1 \right) }{ 2 } \right] ^{ 2 }\)= (136)² = 18496

The find the 8th term we have use the n^th term formula tn = ar^n-1
First term a = 9, common radio r = \(\frac { { t }_{ 2 } }{ { t }_{ 1 } } =\frac { 3 }{ 9 } =\frac { 1 }{ 3 } \)
\({ t }_{ 8 }=9\times \left( \frac { 1 }{ 3 } \right) ^{ 8-1 }=9\times \left( \frac { 1 }{ 3 } \right) ^{ 7 }=\frac { 1 }{ 243 } \)
Therefore the 8th term of the G.P is \(\frac { 1 }{ 243 } \)

Let the consecutive odd integers beginning with 5 be, 5, 7, 9, 11,....
Here a = 5, d = 2
We know that sum of n consecutive integers
\(\mathrm{S}_{\mathrm{n}} =\frac{n}{2}[2 a+(n-1) d] \)
\(480 =\frac{n}{2}[2(5)+(n-1)(2)] \)
\(480 =\frac{2 n}{2}[5+n-1] \)
480 = n(n + 4)
n² + 4n - 480 = 0
(n - 20)(n + 24) = 0
n = 20 (or) -24
Number of terms cannot be negative.
n = 20
20 consecutive odd integers beginning with 5 will sum to 480.

First term a = 5; common difference d = 6.
A.P.is given by., a + d, a + 2d, a + 3d,......
In this case 5, 5 + 6, 5 + 2(6),5 + 3 (6),...
5, 11, 17 ,23,...
The required A.P. is 5, 11,17,23,....

To find a₈, since n = 8 is even, put n = 8 in
\(a_{n}=\frac{n^{2}-1}{n+3}\)
\(a_{8}=\frac{8^{2}-1}{8+3}=\frac{64-1}{11}=\frac{63}{11}\)
To find a₁₅, since n = 15 is odd
put n = 15 in \(a_{n}=\frac{n^{2}}{2 n+1}\)
\(a_{15}=\frac{15^{2}}{2(15)+1}=\frac{225}{30+1}=\frac{225}{31} \)
\(\therefore a_{8}=\frac{63}{11} \text { and } a_{15}=\frac{225}{31} \)

To find a₁₁, Since 11 is odd, we put n = 11 in a_n = n (n + 3)
Thus, the eleventh term a₁₁ = 11(11 + 3) = 154
To find a₁₈, Since 18 is even, we put n = 18 in a_n = n² +1
Thus the eighteenth term a₁₈ = 18² + 1 = 325

In the above sequence the numerators are same and the denominator is increased by 4. 
So the next three terms are
a₅ = \(\frac { 1 }{ 14+4 } =\frac { 1 }{ 18 } \)
a₆ = \(\frac { 1 }{ 18+4 } =\frac { 1 }{ 22 } \)
a₇ = \(\frac { 1 }{ 22+4 } =\frac { 1 }{ 26 } \)