Question

can zero be a term of the ap whose 11th term is 67 and 21st term is 32​

Answer 1

GIVEN :–

• 11th term of A.P. = 67

• 21st term of A.P. = 32

TO FIND :–

• Which term of A.P. will be zero ?

SOLUTION :–

• We know that nth term of A.P. –

$\\ \implies \large { \boxed{\bf T_{n} = a + (n - 1)d}}$

• According to the first condition –

$\\ \implies\bf T_{11} = a + (11- 1)d$

$\\ \implies\bf 67= a + 10d$

$\\ \implies\bf a = 67 - 10d\:\:\:\: \: - - - eq.(1)$

• According to the second condition –

$\\ \implies\bf T_{21} = a + (21- 1)d$

$\\ \implies\bf 32= a +20d$

• Using eq.(1) –

$\\ \implies\bf 32=67 - 10d+20d$

$\\ \implies\bf 32=67 + 10d$

$\\ \implies\bf 32 - 67 = 10d$

$\\ \implies\bf 10d = - 35$

$\\ \implies\bf d = - \dfrac{35}{10}$

$\\ \implies \large{ \boxed{\bf d = - 3.5}}$

• Again using eq.(1) –

$\\ \implies\bf a = 67 - 10( - 3.5)$

$\\ \implies\bf a = 67 + 35$

$\\ \implies \large{ \boxed{\bf a = 102}}$

• Now Let's find which term will be zero.

$\\ \implies\bf 0=102+ (n- 1)( - 3.5)$

$\\ \implies\bf - 102 = (n- 1)( - 3.5)$

$\\ \implies\bf \dfrac{102}{3.5} = (n- 1)$

$\\ \implies\bf n - 1 = 29.14$

$\\ \implies\bf n = 1+ 29.14$

$\\ \implies\bf n=30.14$

▪︎Hence , No term will be zero.