Question

3. Write the first four terms of an A.P. where a= 10 and d=10​

Answer 1

Solution:

Given,

first term ( a ) = 10

common difference ( d ) = 10

We need to find

first 4 terms

Using the formula

tn = a + (n - 1)d

Now , finding

t1 , t2 , t3 , t4 .

t1 = 10 + (1 - 1)d

→ t1 = 10 + 0(d)

→ t1 = 10

t2 = 10 + (2-1)d

→ t2 = 10 + (1)(10)

→ t2 = 10 + 10

→ t2 = 20

t3 = 10 + (3 - 1)d

→ t3 = 10 + 2(10)

→ t3 = 10 + 20

→ t3 = 30

t4 = 10 + (4 - 1)d

→ t4 = 10 + 3(10)

→ t4 = 10 + 30

→ t4 = 40

Hence , first 4 terms are 10 , 20 , 30 , 40

Answer 2

${ \underline{ \blue{ \underline{ \huge{ \tt{QUESTION }}}}}}$




▪ Write the first four terms of an A.P. where a = 10 and d = 10




${ \underline{ \blue{ \huge{ \tt{ \underline{SOLUTION }}}}}}$




${ \rm{{ \underline{ \red{Arithmatic \: Progression}}}}}$



A sequence of numbers which has a common difference between any two consecutive numbers is called an arithmetic progression (A.P.). The example of A.P. is 3,6,9,12,15,18,21, …




✒ The general form of an Arithmetic Progression is a, a + d, a + 2d, a + 3d and so on. 



here,




⛦ a = first term of the series = 10



⛦ d = common difference between the terms = 10




then,




${ \purple{ \sf{first \: term = a = 10}}}$




${ \purple{ \sf{second \: term =( a + d ) }}}\\ \\ { \sf{ \purple{= 10 + 10 = 20}}}$




${ \purple{ \sf{third \: term = (a + 2d)}}} \\ \\ { \purple{ \sf{ = 10 + 2(10)}}} \\ \\ { \purple{ \sf{ = 10 + 20 = 30}}}$



${ \purple{ \sf{ fourth \: term = (a + 3d)}}} \\ \\ { \sf{ \purple{ = 10 + 3(10)}}} \\ \\ { \purple{ \sf{ = 10 + 30 = 40}}}$