Question

5) if the 10th term of an Ap is 52 and 17th term is 20 more than the 13th term, find the 13th term

Arithmetic expressions​

Answer 1

Answer:

a+9d=52

a+16d=20+a+12d

4d=20

a=5

substitute in eq 1

a+9(5)=52

a=7

Answer 2

$\huge\underline\mathbb{\red S\pink{O}\purple{L} \blue{UT} \orange{I}\green{ON :}}$

★ Given that,

↪ 10th term of AP is 52.

• a10 = 52

$\tt\:⟹ a_{10}\:—\:a + 9d = 52....(i)$

↪ 17th term is 20 more than the 13th term.

• a17 = a13 + 20

$\tt\:⟹a + 16d = (a + 12d) + 20$

$\tt\:⟹ 5d = 20$

$\tt\:⟹d = \frac{20}{5}$

$\tt\:⟹d = 5$

$\boxed{∴d = 5}$

• Substitute the value of d in (i).

$\tt\:⟹a + 9(5) = 52$

$\tt\:⟹ a + 45 = 52$

$\tt\:⟹ a = 52- 45$

$\tt\:⟹a = 7$

$\boxed{∴a = 7}$

$\tt\:⟹a_{13}=a + 12d$

• Substitute the values.

$\tt\:⟹ a_{13} = 7 + 12(5)$

$\tt\:⟹ a_{13}=7 + 60$

$\tt\:⟹ a_{13}= 67$

$\underline{\boxed{\bf{\purple{∴ The \: value \: of \: a_{13} \: = \: 67}}}}$