Question

ㅤㅤㅤㅤ ★ Maths Query ★


1: Find the $\sf{20^{th}}$ term from the last term of the AP : $\boxed{\sf{\purple{ 3 \:,\: 8 \:,\: 13 \:,\: . . . . . . \:,\: 253 }}}$

2: Rihan started work at 1995 at an annual salary of Rs 5000 & received an increment of Rs 200 each year . In which year did his income reach Rs 7000 ?

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Topic : Arithmetic Progressions​

Answer 1

Answer:

For the given A.P.,

First term(a)=3

Last term(t

n

)=253

Common Difference(d)=8−3=5

∵t

n

=a+(n−1)d

⇒253=3+(n−1)×5

⇒253−3=(n−1)×5

⇒250÷5=n−1

⇒50+1=n

∴n=51

In 51 terms, 20

th

term from the last term will be 51−20+1=32

th

∴t

32

=a+(32−1)d=3+31×5=3+155=158

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Answer 2

Answer:

158

11 year

Step-by-step explanation:

(i) Here, first term = a = 3

common difference = d = 8 - 3 = 5

Last term = l = 253

Using nth term = a + (n - 1)d, we get

=> 253 = 3 + (n - 1)5

=> 51 = n

We are asked for the 20th term from the end which is (51 - 20 + 1)th term from the starting.

From starting, (51 - 19)th term = 32th term(from start)

=> 32th term = 3 + (32 - 1)5 = 158

It can also be solved as:

5 is the common difference from start, it means, -5 is the common difference from end. From end, first term is 253. Thus

=> 20th term = 253 + (20 - 1)(-5) = 158

(ii) At the starting:

Salary(constant) = a = 5000

Increment(comm. diff.) = d = 200

Last salary(last term) = l = 7000

Let the no. of year be 'n'.

=> l = a + (n - 1)d

=> 7000 = 5000 + (n - 1)200

=> 11 = n

It requires 11 year.