If a1- a2- a15 are in A-P- and a1-a8-a15- 15- then a2-a3-a8-a13-a14 equals

A_1+a_8+a_15= 3a_1 + 21d = 15 ⇒ a_1 + 7d = 5 Now, a_2+a_3+a_8+a_13+a_14 = 5a_1 + 35d 5(a_1+7d)=5× 5=25 ...

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Selecting Consecutive Terms in A.P

If a1, a2,···...

Question

If $a_{1}, a_{2}, \cdot \cdot \cdot, a_{15}$ are in A.P. and $a_{1} + a_{8} + a_{15} = 15$ , then $a_{2} + a_{3} + a_{8} + a_{13} + a_{14}$ equals

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a_{1}, a_{2}, a_{3}, . . ., a_{15}

a_{1} + a_{8} + a_{15} = 15

a_{2} + a_{3} + a_{8} + a_{13} + a_{14} =

If $a_{1}, a_{2}, \cdot \cdot \cdot, a_{15}$ are in A.P. and $a_{1} + a_{8} + a_{15} = 15$ , then $a_{2} + a_{3} + a_{8} + a_{13} + a_{14}$ equals

a_{1}, a_{2}, a_{3} . . . . . . ., a_{15} a r e i n A . P a n d a_{1} + a_{8} + a_{15} = 15 t h e n a_{2} + a_{3} + a_{8} + a_{13} + a_{14}

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

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

a_{1} + a_{2} + a_{3} + . . . + a_{24}

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

A . P .

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

a_{1} + a_{2} + a_{3} + . . . + a_{23} + a_{24}

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