The 9thterm of an AP is 449 and 448th term is 9. Find the 458th term of the A.P. About the author Margaret
Answer:– Given: 9th term of an AP (a₉) = 449 448th term of an AP (a₄₄₈) = 9. We know that; nth term of an AP (aₙ) = a + (n – 1)d Hence; a₉ = a + (9 – 1)d ⟹ 448 = a + 8d ⟹ 448 – 8d = a — equation (1). Similarly; ⟹ a + (448 – 1)d = 9 ⟹ a + 447d = 9 Substitute the value of a from equation (1). ⟹ 448 – 8d + 447d = 9 ⟹ 439d = 9 – 448 ⟹ 439d = – 439 ⟹ d = – 439/439 ⟹ d = – 1 Substitute the value of d in equation (1). ⟹ a = 448 – 8d ⟹ a = 448 – 8( – 1) ⟹ a = 448 + 8 ⟹ a = 456 Now; 458th term of the AP (a₄₅₈) = a + 457d ⟹ a₄₅₈ = 456 + 457( – 1) ⟹ a₄₅₈ = 456 – 457 ⟹ a₄₅₈ = – 1 ∴ The 458th term of the given AP is – 1. Reply
Answer:–
Given:
9th term of an AP (a₉) = 449
448th term of an AP (a₄₄₈) = 9.
We know that;
nth term of an AP (aₙ) = a + (n – 1)d
Hence;
a₉ = a + (9 – 1)d
⟹ 448 = a + 8d
⟹ 448 – 8d = a — equation (1).
Similarly;
⟹ a + (448 – 1)d = 9
⟹ a + 447d = 9
Substitute the value of a from equation (1).
⟹ 448 – 8d + 447d = 9
⟹ 439d = 9 – 448
⟹ 439d = – 439
⟹ d = – 439/439
⟹ d = – 1
Substitute the value of d in equation (1).
⟹ a = 448 – 8d
⟹ a = 448 – 8( – 1)
⟹ a = 448 + 8
⟹ a = 456
Now;
458th term of the AP (a₄₅₈) = a + 457d
⟹ a₄₅₈ = 456 + 457( – 1)
⟹ a₄₅₈ = 456 – 457
⟹ a₄₅₈ = – 1
∴ The 458th term of the given AP is – 1.