Question

If five times the fifth term of an A.P is equal to 8times its eigth term show that its 13th is zero

Answer 1

Hello Harshit
let the first term of the AP be a, and common difference be d.
Then nth term,An=a+(n-1)d
Now according to question,
5 x 5th term= 8 x 8th term
or,5 x (a+(5-1)d)=8x (a+(8-1)d)
or,5 x(a+4d)=8 x (a+7d)
or,5a+20d=8a+56d
or,5a-8a=36d
or,-3a=36d
or,a=-12d. (1)
Now 13 th term will be
a+(13-1)d
or,a+12d
or,a-a=0. (from 1)

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: QUESTION \: \: } \mid}}}}}$

$\: \: \: \: \: \: \: \: \: \: \text{If 5 times the 5th term of an AP is equal } \\ \text{to 8 times its 8th term, show that the 13th} \\ \text{ term is zero.}</p><p>$

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\underline{ \mathfrak{ \: \: Given, \: \: }} \\\ \\ \mathtt{ \rightsquigarrow 5 \: times \: \: of \: \: the \: \: 5th \: \: term = 8th \: \:times} \\ \mathtt{ \: \: \: \: \:\: \: \: \: \:\: \: \: \: \: \:\: \: \: \: \: \: \: \: \:\: \: \:\: \: \:\: \: \: \: of \: \: the \: \: 8th \: \: term} \\ \\ \\ \underline{ \mathfrak{ \: \: </p><p>To \: show : \mapsto \: \: }} \\ \\ \mathtt{ \rightsquigarrow \: 13th \: \: term \: \: is \: \: equal \: \: to \: \: zero. \: }$

$\underline{ \mathfrak{ \: \: Suppose, \: \: }} \\ \\ \mathtt{ \rightsquigarrow \: First \: \: term \: \: of \: \: the \: \: A.P. = a} \\ \mathtt{ \rightsquigarrow \: Difference = d}$

$\underline{ \bold{ \: A.T.Q., \: }} \\ \\ \: \: \: \: \: \: \mathtt{ 5 \times T_5 = 8 \times T_8 }\\ \\ \mathtt{\Longrightarrow \: 5[ a + (5-1)d]=8 [ a + (8-1)d]} \\ \\ \mathtt{\Longrightarrow 5(a + 4d)= 8 (a+7d) }\\ \\ \mathtt{ \Longrightarrow 5a + 20 \: d = 8a + 56 \: d} \\ \\ \mathtt{ \Longrightarrow 5a - 8a = 56 \: d - 20 \: d}\\ \\ \mathtt{\Longrightarrow - 3a = 36 \: d}\\ \\ \mathtt{\Longrightarrow a = \frac{36 \: d}{ - 3}} \\ \\ \: \: \: \mathtt{ \therefore \: \: \underline{ \: a = - 12 \: d \: }}</p><p></p><p>$

$\underline{\mathfrak{ \: Now \: }} \\ \\ \mathtt {13th \: \: term, \: T_{13 }= a + [(13-1)d] } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{ = a + 12 \: d} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{ = ( - 12 \: d) + 12 \: d } \\ \mathtt{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{[As \: \: a = - 12 \: d]}} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\mathtt{ = - 12 \: d + 12 \: d} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{ = 0} \\ \\ \\ \huge{\mathtt{ \therefore \: \: \underline{ \: \: 13th \: \: term \: = 0 \: \: }} }\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: </p><p> \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{ \underline{ \: Hence \: \: proved. \: \: }}$