Question

If five times the 5th term of an AP is equal to 8times it's 8th term show that it's 13th term zero

Answer 1

Hey there !!

Given :-

→ 5( a₅ ) = 8( a₈ ) .

To prove :-

→ a₁₃ = 0 . .

Solution :-

We have,

⇒ 5( a₅ ) = 8( a₈ ) .

⇒ 5( a + 4d ) = 8 ( a + 7d ) .

⇒ 5a + 20d = 8a + 56d .

⇒ 20d - 56d = 8a - 5a .

⇒ - 36d = 3a .

⇒ 3a = -36d .

⇒ 3a + 36d = 0 .

⇒ 3( a + 12d ) = 0 .

⇒ a + 12d = 0/3 .

⇒ a + 12d = 0 .

⇒ a + ( 13 - 1 )d = 0 .

∴ a₁₃ = 0 .

Hence, 13th term of the AP is 0 .

THANKS.

#BeBrainly .

Answer 2

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

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

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

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

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

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

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