Question

if 20th term and 30th term of Arthematic progression are 121 and 181 respectively find the 40th term of Arthematic progression​

Answer 1

✬ 40th Term = 241 ✬

Step-by-step explanation:

Given:

• 20th and 30th term of an AP is 121 and 181 respectively.

To Find:

• What is the 40th term of AP ?

Solution: As we know that an AP series is given by

★ a + (n – 1)d ★

• a = first term

• n = number of terms

• d = common difference

A/q

• 20th term is 121.

➙ a + (20 – 1)d = 121

➙ a + 19d = 121

➙ a = 121 – 19dㅤㅤㅤㅤㅤeqⁿ i

Now ,

• 30th term is 181

➙ a + (30 – 1)d = 181

➙ a + 29d = 181

➙ 121 – 19d + 29d = 181 ㅤㅤㅤfrom eqⁿ i

➙ 10d = 181 – 121

➙ d = 60/10 = 6

So the common difference of AP is 6. Now putting the value of d in eqⁿ 1.

➮ a = 121 – 19 × 6

➮ a = 121 – 114

➮ a = 7

So the first term of AP is 7.

∴ 40th term will be

$\implies{\rm }$ a + (40 – 1)d

$\implies{\rm }$ 7 + 39 × 6

$\implies{\rm }$ 7 + 234

$\implies{\rm }$ 241

Hence, the 40th term of AP will be 241.

Answer 2

Answer:

241

Step-by-step explanation:

20th term = 121

a + 19d = 121 equation 1

30th term = 181

a + 29 d = 181 equation(2)

solving equation 1 and 2 we get

d = 6

from equation 1

a = 121 - 19x6

a= 7

therefore

30th term = 7 + 29x6

= 241