Question

In an AP, if a = 4, n = 7 and = 4, then the value of ‘d’

a. 0

b. 1

c. 3

d. 2​

Answer 1

Answer:

$\huge{\pink{\boxed{\green{\sf{d=0}}}}}$

Step-by-step explanation:

$\huge{\pink{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}$

Correction in question :

□a7=4

$\: \: { \orange{ \underline{given \ratio}}} \\ {\green{a = 4}} \\ {\green{n = 7}} \\ {\green{an = 4}} \\ \\ \: \: \: \: \: \: \: \: \: \: { \blue{ \underline{to \: find \ratio}}} \\ {\red{common \: difference(c.d) = }}$

☆ We know the formula to find differnce if first term , number of term and last term given :

$\to \: an = a + (n - 1)d \\ \to 4 = 4 + (7 - 1)d \\ \to \: 4 - 4 = 6d \\ \to \: 6d = 0 \\ \to {\boxed{d = 0}}$

$\huge{\pink{\boxed{\green{\sf{d=0}}}}}$

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□Some more information about A.P

${\underline{note - }} \\ \to \: value \: of \: n \: cannot \: be \: in \: decimal \: negative \: form \\ {\underline{ formula - }} \\ \to \: an = a + (n -1 )d \\ \to \: sn = \frac{n}{2} (2a + (n - 1)d) \\ where \ratio\\ \to \: an = last \: term \\ \to \: a = first \: term \\ \to \: d \: = common \: difference \\ \to \: n = number \: of \: terms \\ \to \: sn = sum \: of \: nth \: term$

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Answer 2

Answer:

(a) 0

Step-by-step explanation:

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