In an AP $t_{1} + t_{7} = 18, t_{4} + t_{10} = 54$ find twenty value of $520$

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Solution

Let $t_{1} = a$ and common difference be $d$

Given that $\Rightarrow t_{1} + t_{7} = 18$

$\Rightarrow a + (a + 6 d) = 18$

$\Rightarrow 2 a + 6 d = 18$

$\Rightarrow a + 3 d = 9$ ... $(1)$

Given that $\Rightarrow t_{4} + t_{10} = 54$

$\Rightarrow (a + 3 d) + (a + 9 d) = 54$

$\Rightarrow 2 a + 12 d = 54$

$\Rightarrow a + 6 d = 27$ .... $(2)$

On solving we get,

$\Rightarrow a = - 9$ and $d = 6$

$\Rightarrow t_{20} = a + 19 d$

$\Rightarrow t_{20} = - 9 + 19 (6)$

$\Rightarrow t_{20} = 105$

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Similar questions

(a) In a H.P, find T₁₉

(b) In a H.P, find T₇ and T₁₂

t_{54}

t_{4} = 64

t_{10}

t_{54}

- 61

t_{4} = 64,

t_{10} .

T_{5} : T_{10} = 32 : 1

T_{7} = \frac{1}{32}

T_{1} * T_{2} * T_{7}

T_{7} = 9

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