| Numbers\[{{a}_{1}},\]\[{{a}_{2}},\]\[{{a}_{3}},\]\[{{a}_{4}},\]\[{{a}_{5}},...,\]\[{{a}_{24}}\]with common difference = 10, are in arithmetic progression and\[{{a}_{1}}+{{a}_{5}}+{{a}_{10}}+{{a}_{15}}+{{a}_{20}}+{{a}_{25}}=225.\] The value of \[{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}}+{{a}_{5}}+...+{{a}_{23}}+{{a}_{24}}\]is |
A) 525
B) 725
C) 860
D) 900
Correct Answer: C
Solution :
| As \[{{a}_{1}},\]\[{{a}_{2}},\]\[{{a}_{3}},\] \[{{a}_{4}},....,\]\[{{a}_{24}}\]are in AP. |
| So, let \[{{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}={{a}_{4}}-{{a}_{3}}=....=d\] |
| \[\therefore \] \[{{a}_{2}}={{a}_{1}}+d\]\[\Rightarrow \]\[{{a}_{3}}={{a}_{1}}+2d\] |
| \[{{a}_{4}}={{a}_{1}}+3d\]\[\Rightarrow \]\[{{a}_{24}}={{a}_{1}}+23d\] |
| According to the question, |
| \[{{a}_{1}}+{{a}_{5}}+{{a}_{10}}+{{a}_{15}}+{{a}_{20}}+{{a}_{25}}=225\] |
| \[{{a}_{1}}({{a}_{1}}+4d)+({{a}_{1}}+9d)+({{a}_{1}}+14d)\] \[+\,\,({{a}_{1}}+19d)+({{a}_{1}}+24d)=225\] |
| \[\Rightarrow \]\[6{{a}_{1}}+70d=225\] (i) |
| Now, \[{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}}+...+{{a}_{24}}\] |
| \[={{a}_{1}}+({{a}_{1}}+d)+({{a}_{1}}+2d)+({{a}_{1}}+3d)\] |
| \[+...+({{a}_{1}}+23d)\] |
| \[=24{{a}_{1}}+d\,\,(1+2+3+4+5+...+23)\] |
| \[=24{{a}_{1}}+d\left( \frac{23\times 24}{2} \right)\] |
| \[=24{{a}_{1}}+276\,\,d=4\,\,(6{{a}_{1}}+69d)\] |
| \[=4\,\,(225-70d+69d)\] [from Eq. (i)] |
| \[=900-4d\] |
| \[=900-40=860\]\[[\because d=10,\text{given }\!\!]\!\!\text{ }\] |
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