If the ratio of the sum of the first n terms of two A.Ps is (7n+1):(4n+27) , then find the ratio of their 9th terms.
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Arithmetic progression. AP.
To find ratio of 9th terms given ratio of sum of n terms.
![S1_n=\frac{n}{2}[2a_1+(n-1)d_1]\\S2_n=\frac{n}{2}[2a_2+(n-1)d_2]\\\\\frac{S1_n}{S2_n}=\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{7n+1}{4n+27}=\frac{8+(n-1)7}{31+(n-1)4}\\\\So \: \: a_1=4k, \: a_2=31 k, \: \: d_1=7k, \: \: d_2=4k, \ \ for \: some \: k.\\\\LHS=\frac{T1_9}{T2_9}=\frac{a_1+8d_1}{a_2+8d_2}=\frac{4k+8*7k}{31k+8*4k}\\\\=\frac{20}{21}](https://tex.z-dn.net/?f=S1_n%3D%5Cfrac%7Bn%7D%7B2%7D%5B2a_1%2B%28n-1%29d_1%5D%5C%5CS2_n%3D%5Cfrac%7Bn%7D%7B2%7D%5B2a_2%2B%28n-1%29d_2%5D%5C%5C%5C%5C%5Cfrac%7BS1_n%7D%7BS2_n%7D%3D%5Cfrac%7B2a_1%2B%28n-1%29d_1%7D%7B2a_2%2B%28n-1%29d_2%7D%3D%5Cfrac%7B7n%2B1%7D%7B4n%2B27%7D%3D%5Cfrac%7B8%2B%28n-1%297%7D%7B31%2B%28n-1%294%7D%5C%5C%5C%5CSo+%5C%3A+%5C%3A+a_1%3D4k%2C+%5C%3A+a_2%3D31+k%2C+%5C%3A+%5C%3A+d_1%3D7k%2C+%5C%3A+%5C%3A+d_2%3D4k%2C+%5C+%5C+for+%5C%3A+some+%5C%3A+k.%5C%5C%5C%5CLHS%3D%5Cfrac%7BT1_9%7D%7BT2_9%7D%3D%5Cfrac%7Ba_1%2B8d_1%7D%7Ba_2%2B8d_2%7D%3D%5Cfrac%7B4k%2B8%2A7k%7D%7B31k%2B8%2A4k%7D%5C%5C%5C%5C%3D%5Cfrac%7B20%7D%7B21%7D "S1_n=\frac{n}{2}[2a_1+(n-1)d_1]\\S2_n=\frac{n}{2}[2a_2+(n-1)d_2]\\\\\frac{S1_n}{S2_n}=\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{7n+1}{4n+27}=\frac{8+(n-1)7}{31+(n-1)4}\\\\So \: \: a_1=4k, \: a_2=31 k, \: \: d_1=7k, \: \: d_2=4k, \ \ for \: some \: k.\\\\LHS=\frac{T1_9}{T2_9}=\frac{a_1+8d_1}{a_2+8d_2}=\frac{4k+8*7k}{31k+8*4k}\\\\=\frac{20}{21}")
Answer is 20/21.
To find ratio of 9th terms given ratio of sum of n terms.
Answer is 20/21.
kvnmurty:
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answer is wrong
answer is 24/19
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Hey mate..
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Please see the given pic attentively to figure the solution out.
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Please see the given pic attentively to figure the solution out.
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