Let there be an A.P. with first term 'a', common difference 'd'.

Solution:

(i) Here, we have an A.P. whose nth term (an), first term (a) and common difference (d) are given. We need to find the number of terms (n) and the sum of first n terms (Sn).

Here,

First term (a) = 5

Last term () = 50

Common difference (d) = 3

So here we will find the value of n using the formula, 

So, substituting the values in the above mentioned formula

$50=5+(n-1) 3$

$50=5+3 n-3$

$50=2+3 n$

$3 n=50-2$

Further simplifying for n,

$3 n=48$

$n=\frac{48}{3}$

 

$n=16$

Now, here we can find the sum of the n terms of the given A.P., using the formula,

Further simplifying for n,

$3 n=48$

$n=\frac{48}{3}$

$n=16$

Now, here we can find the sum of the n terms of the given A.P., using the formula,

$S_{n}=\left(\frac{n}{2}\right)(a+l)$

Where, a = the first term

l = the last term

So, for the given A.P, on substituting the values in the formula for the sum of n terms of an A.P., we get,

$S_{16}=\left(\frac{16}{2}\right)[5+50]$

$=8(55)$

$=440$

Therefore, for the given A.P $n=16$ and $S_{16}=440$

(ii) Here, we have an A.P. whose nth term (an), sum of first n terms (Sn) and common difference (d) are given. We need to find the number of terms (n) and the first term (a).

Here,

Last term $\left(a_{e}\right)=4$

Common difference $(d)=2$

Sum of $n$ terms $\left(S_{n}\right)=-14$

So here we will find the value of $n$ using the formula, $a_{n}=a+(n-1) d$

So, substituting the values in the above mentioned formula

$4=a+(n-1) 2$

$4=a+2 n-2$

$4+2=a+2 n$

$n=\frac{6-a}{2}$ .......(1)

Now, here the sum of the n terms is given by the formula,

$S_{n}=\left(\frac{n}{2}\right)(a+l)$

Where, a = the first term

l = the last term

So, for the given A.P, on substituting the values in the formula for the sum of n terms of an A.P., we get,

$-14=\left(\frac{n}{2}\right)[a+4]$

$-14(2)=n(a+4)$

$n=\frac{-28}{a+4}$ .........(2)

Equating (1) and (2), we get,

$\frac{6-a}{2}=\frac{-28}{a+4}$

$(6-a)(a+4)=-28(2)$

$6 a-a^{2}+24-4 a=-56$

$-a^{2}+2 a+24+56=0$

So, we get the following quadratic equation,

$-a^{2}+2 a+80=0$

$a^{2}-2 a-80=0$

Further, solving it for a by splitting the middle term,

$a^{2}-2 a-80=0$

$a^{2}-10 a+8 a-80=0$

$a(a-10)+8(a-10)=0$

 

$(a-10)(a+8)=0$

So, we get,

$a-10=0$

$a=10$

Or

$a+8=0$

$a=-8$

Substituting,  in (1),

$n=\frac{6-10}{2}$

$n=\frac{-4}{2}$

$n=-2$

Here, we get $n$ as negative, which is not possible. So, we take $a=-8$,

$n=\frac{6-(-8)}{2}$

$n=\frac{6+8}{2}$

$n=\frac{14}{2}$

$n=7$

Therefore, for the given A.P $n=7$ and $a=-8$

(iii) Here, we have an A.P. whose first term (a), sum of first n terms (Sn) and the number of terms (n) are given. We need to find common difference (d).

Here,

First term () = 3

Sum of n terms (Sn) = 192

Number of terms (n) = 8

So here we will find the value of n using the formula, 

So, to find the common difference of this A.P., we use the following formula for the sum

of n terms of an A.P

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 8, we get,

$S_{8}=\frac{8}{2}[2(3)+(8-1)(d)]$

$192=4[6+(7)(d)]$

$192=24+28 d$

$28 d=192-24$

Further solving for d,

$d=\frac{168}{28}$

$d=6$

Therefore, the common difference of the given A.P. is.

(iv) Here, we have an A.P. whose nth term (an), sum of first n terms (Sn) and the number of terms (n) are given. We need to find first term (a).

Here,

Last term () = 28

Sum of n terms (Sn) = 144

Number of terms (n) = 9

Now,

$a_{9}=a+8 d$

$28=a+8 d$.............(1)

Also, using the following formula for the sum of n terms of an A.P

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 9, we get,

$S_{8}=\frac{9}{2}[2 a+(9-1)(d)]$

$144(2)=9[2 a+8 d]$

$288=18 a+72 d$.........(2)

Multiplying (1) by 9, we get

$9 a+72 d=252$ ........(3)

Further, subtracting (3) from (2), we get

$9 a=36$

$a=\frac{36}{9}$

$a=4$

Therefore, the first term of the given A.P. is $a=4$.

(v) Here, we have an A.P. whose nth term (an), sum of first n terms (Sn) and first term (a) are given. We need to find the number of terms (n) and the common difference (d).

Here,

First term () = 8

Last term () = 62

Sum of n terms (Sn) = 210

Now, here the sum of the n terms is given by the formula,

$S_{n}=\left(\frac{n}{2}\right)(a+l)$

Where, a = the first term

l = the last term

So, for the given A.P, on substituting the values in the formula for the sum of n terms of an A.P., we get,

$210=\left(\frac{n}{2}\right)[8+62]$

$210(2)=n(70)$

$n=\frac{420}{70}$

$n=6$

Also, here we will find the value of d using the formula,

$a_{n}=a+(n-1) d$

So, substituting the values in the above mentioned formula

$62=8+(6-1) d$

$62-8=(5) d$

$\frac{54}{5}=d$

$d=\frac{54}{5}$

Therefore, for the given A.P $n=6$ and $d=\frac{54}{5}$

(vi) Here, we have an A.P. whose first term (a), common difference (d) and sum of first n terms are given. We need to find the number of terms (n) and the nth term (an).

Here,

First term (a) = 2

Sum of first nth terms () = 90

Common difference (d) = 8

So, to find the number of terms (n) of this A.P., we use the following formula for the sum

of n terms of an A.P

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 8, we get,

$S_{n}=\frac{n}{2}[2(2)+(n-1)(8)]$

$90=\frac{n}{2}[4+8 n-8]$

$90(2)=n[8 n-4]$

$180=8 n^{2}-4 n$

Further solving the above quadratic equation,

$8 n^{2}-4 n-180=0$

$2 n^{2}-n-45=0$

Further solving for n,

$2 n^{2}-10 n+9 n-45=0$

$2 n(n-5)+9(n-5)=0$

 

$(2 n+9)(n-5)=0$

Now,

$2 n+9=0$

$2 n=-9$

$n=-\frac{9}{2}$

Also,

$n-5=0$

$n=5$

Since n cannot be a fraction

Thus, n = 5

Also, we will find the value of the nth term (an) using the formula, 

So, substituting the values in the above mentioned formula

$a_{n}=2+(5-1) 8$

$a_{n}=2+(4)(8)$

$a_{n}=2+32$

$a_{n}=34$

Therefore, for the given A.P $n=5$ and $\mathrm{a}_{n}=34$.

(vii)

$a_{k}=S_{k}-S_{k-1}$

$\Rightarrow 164=\left(3 k^{2}+5 k\right)-\left(3(k-1)^{2}+5(k-1)\right)$

$\Rightarrow 164=3 k^{2}+5 k-3 k^{2}+6 k-3-5 k+5$

$\Rightarrow 164=6 k+2$

$\Rightarrow 6 k=162$

$\Rightarrow k=27$