For an A-P- a1-a2-an and a1-5-2-a10-16- If sum of n terms is 110 then n equals

The correct option is D 11 a_1=52 Let the common difference be d.Then a_10=a_1+9d 16=52+9d 272=9d d=32 Hence S_n=n2(2(52)+(n ...

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Formula for Sum of n Terms of an AP

For an A.P. ...

Question

For an A.P. $a_{1}, a_{2}, . . . a_{n}$ and $a_{1} = \frac{5}{2}, a_{10} = 16$ .
If sum of $n$ terms is $110$ then $n$ equals

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a_{1}, a_{2}, a_{3}, \dots

\frac{a_{1} + a_{2} + \dots + a_{10}}{a_{1} + a_{2} + \dots + a_{p}} = \frac{100}{p^{2}},

p \neq 10,

\frac{a_{11}}{a_{10}}

a_{1}, a_{2}, a_{3}, . . . . . . . . . ., a_{n}

A . P .

(1^{2} - a_{1}) + (2^{2} - a_{2}) + (3^{2} - a_{3}) + . . . . . . . + (n^{2} - a_{n}) = \frac{(n - 1) n (n + 1)}{3}

(\frac{a_{5} + a_{3} - a_{2}}{6})

a_{1}, a_{2}, a_{3} \dots

a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225

A . P .

a_{1}, a_{2}, a_{3}, . . . . . ., a_{n}, . . . . .

a_{1} + a_{3} + a_{5} = - 12

a_{1} a_{2} a_{3} = 8

a_{2} + a_{4} + a_{6}

a_{1}, a_{2}, a_{3},

a_{n}

a_{4} - a_{7} + a_{10} = m

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