Question

th
1.
The nth term of an A.P. is given by an = 3 + 4n. The common difference is
(a) 7
(b)3
(c) 4
(d) 1​

Answer 1

$\red{ \underline{answer}} \\ \\ \implies {common \: \: difference \: \: = \: \: 4} \\ \\ \implies \green{ \underline{to \: \: find}} \\ \\ \implies {common \: \: difference} \\ \\ \implies \orange{ \underline{solution}} \\ \\ \implies {nth \: \: terms \: \: of \: \: ap \: } \\ \\ \implies {an = 3 + 4n \: \: (given)} \\ \\ \implies {put \: \: n \: = 1 = 7} \\ \\ \implies {put \: \: n \: = 2 = 3 + 8 = 11} \\ \\ \implies {put \: \: n \: = 3 = 3 + 12 = 15} \\ \\ \implies {put \: \: n \: = 4 =3 + 16 = 19 } \\ \\ \implies {ap \: \: = \: \: 7 + 11 + 15 + 19....... \: n \: terms} \\ \\ \implies {first \: \: term \: \: = a \: = 7} \\ \\ \implies {common \: \: difference \: = d = 11 - 7 = 4} \\ \\ \implies \pink{ \underline{verification}} \\ \\ \implies { \boxed{an \: = \: a \: + (n - 1)d}} \\ \\ \implies {an \: = 7 + (n - 1)4} \\ \\ \implies {7 + 4n - 4} \\ \\ \implies {3 + 4n} \\ \\ \implies \green \therefore \green{ \boxed{verified}} \\ \\ \implies \orange{ \underline{related \: \: formula}} \\ \\ \implies { \boxed{an \: = a + (n - 1)d}} \\ \\ \implies { \boxed{sn \: = \frac{n}{2}(2a + (n - 1)d }}$

Answer 2

QUESTION:

The nth term of an A.P. is given by an = 3 + 4n. The common difference is

(a) 7

(b)3

(c) 4

(d) 1

ANSWER:

The nth term of an AP is given by;

$an = 3 + 4n$

If we put the value 1 that is a1 then we will get the first term;

$a(1) = 3 + 4 \times 1 \\ a(1) = 3 + 4 \\ a(1) = 7$

$\huge\orange{first \: term = 7}$

If we put the value 2 that is a2 then we will get the second term ;

$a(2) = 3 + 4 \times 2 \\ a(2) = 3 + 8 \\ a(2) = 11$

$\huge\pink {second \: term = 11}$

now,

we know that;

$\blue {second \: term - first \: term = common \: differnce}$

$\red{a2 - a1 = d}$

so,

$d = 11 - 7 \\ d = 4$

$\huge\green {common \: differnce = 4}$

your answer is option c ✔✔.

plz Mark as brainliest.