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ICSE Class 10 Mathematics Question 13 of 21

Solved 2024 Question Paper ICSE Class 10 Mathematics — Question 7

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7
Question

Question 6(ii)

The nth term of an Arithmetic Progression (A.P.) is given by the relation Tn = 6(7 - n). Find ∶

(a) its first term and common difference

(b) sum of its first 25 terms

Answer

(a) Tn = 6(7 - n)

Substituting n = 1, we get :

T1 = 6(7 - 1) = 6 × 6 = 36.

Substituting n = 2, we get :

T2 = 6(7 - 2) = 6 × 5 = 30.

Common difference (d) = T2 - T1

= 30 - 36 = -6.

Hence, common difference (d) = -6 and first term = 36.

(b) By formula,

Sum upto n terms = n2[2a+(n−1)d]\dfrac{n}{2}[2a + (n - 1)d]

Sum upto 25 terms=252[2×36+(25−1)×−6]=12.5×[72+24×−6]=12.5×[72−144]=12.5×−72=−900.\text{Sum upto 25 terms} = \dfrac{25}{2}[2 \times 36 + (25 - 1) \times -6] \\[1em] = 12.5 \times [72 + 24 \times -6] \\[1em] = 12.5 \times [72 - 144] \\[1em] = 12.5 \times -72 \\[1em] = -900.

Hence, sum of its first 25 terms = -900.