Q) -11, -7, -3, ………….,49, 53 are the terms of a progression. Answer the following:

(a) What is the type of progression?

(b) How many terms are there in all?

(c) Calculate the value of middle most term.

ICSE Specimen Question Paper 2026

Ans:

a) Type of progression:

Step 1: given progression is: -11, -7, -3, ………….,49, 53

Here, we can see that the first term of the progression, a = – 11

and the difference between any 2 consecutive terms is 4

(-7) – (-11) = -7 + 11 = 4

53 – 49 = 4

Since the difference is common across the progression, this is an Arithmetic Progression (AP).

(b) Number of terms in AP:

Step 2: Let’s consider there are n terms in all and hence, the value of n^th term (last term of AP) is 53 (given)

Now, we know that n^th term of an AP, T_n  =  a + (n – 1) d

Therefore, 53 = (- 11)  + (n – 1) (4)

∴ 53 + 11 = 4 (n – 1)

∴ 64 = 4 (n – 1)

∴ n – 1 = = 16

∴ n = 16 + 1 = 17

Therefore, there are 17 terms in all in the given AP.

(c) Value of middle most term:

Step 3: Since, there are 17 terms in the given AP, its middle most term will be 9th term, where 8 terms will be on its left hand side and 8 terms will be on the right hand side.

Let’s find the value of this 9th term of the AP, whose first term, a is – 11 and common difference, d is 4

Since n^th term of an AP, T_n  =  a + (n – 1) d

∴ T₉  =  (- 11) + (9 – 1) (4) = -11 + 8 x 4

∴ T₉  =  = -11 + 32 = 21

Therefore, the 9th term is middle most term of the AP and its value is 21.

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