Q3 of 50 Page 21

The 5th term of an arithmetic sequence is 38 and the 9th term is 66. What is its 25th term?

Given:


a5 = 38 and,


a9 = 66


We know, nth term is given as,


[an = a + (n-1)× d]


⇒ a5 = a + 4d ..(1)


And, a9 = a + 8d …(2)


Solving eq(1) and eq(2), we get,


a9 = a5 + 4d


⇒ 66 = 38 + 4d


⇒ .


Now, from eq (1),


a5 = a + 4d


⇒ 38 = a + 4(7)


⇒ a = (38-(4×7))


⇒ a = 38 – 28


⇒ a = 10


And,


25th term, a25 = a + 24d


= 10 + (24×7)


= 178


Hence, 25th term is 178


More from this chapter

All 50 →
1

In each of the arithmetic sequences below, some terms are missing and their positions are marked with ∘. Find them.

i) 24, 42, ∘, ∘, … ii) ∘, 24, 42, ∘, …


iii) ∘, ∘, 24, 42, … iv) 24, ∘, 42, ∘, …


v) ∘, 24, ∘, 42, … vi) 24, ∘, ∘, 42, …

 2

The terms in two positions of some arithmetic sequences are given below. Write the first five terms of each:

i) 3rd term 34


6th term 67


ii) 3rd term 43


6th term 76


iii) 3rd term 2


5th term 3


iv) 4th term 2


7th term 3


v) 2nd term 5


5th term 2

 4

Is 101 a term of the arithmetic sequence 13, 24,35,…? What about 1001?

 5

How many three-digit numbers are there, which leave a remainder 3 on division by 7?